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MATHEMATICS 


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we Of 
» 


OF THE 


INTEGRAL CALCULUS, 


WITH A 


KEY TO THE SOLUTION OF DIFFERENTIAL 
EQUATIONS, AND A SHORT TABLE 
OF INTEGRALS. 


BY 


WILLIAM ELWOOD BYERLY, PuD., 


PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY. a 


SECOND EDITION, REVISED AND ENLARGED. 


BOSTON, U.S.A.: 
PUBLISHED BY GINN AND COMPANY. 
1889. 


Entered according to Act of Congress, in the year 1888, by 
WILLIAM ELWOOD BYERLY, 
in the Office of the Librarian of Congress at Washington. 


ALL Rients RESERVED. 


TyYPoGRAPHY BY J. S. CusHine & Co., Boston, U.S.A. 


PRESSWORK BY GINN & Co., Boston, U.S.A. 


+ 4a2t- Sale £5 
Rasy ete Paper tests | 


PREFACE. LE 


Tue following volume is a sequel to my treatise on the 
Differential Calculus, and, like that, is written as a text-book. 
The last chapter, however, a Key to the Solution of Differential 
Equations, may prove of service to working mathematicians. 

I have used freely the works of Bertrand, Benjamin Peirce, 
Todhunter, and Boole; and I am much indebted to Professor 
J. M. Peirce for criticisms and suggestions. 

I refer constantly to my work on the Differential Calculus 
as Volume I.; and for the sake of convenience I have added 
Chapter V. of that book, which treats of Integration, as an 


appendix to the present volume. 
W. E. BYERLY. 
CAMBRIDGE, 1881. 


PREFACE TO SECOND EDITION. 


In enlarging my Integral Calculus I have used freely 
Schlomilch’s ‘‘ Compendium der Hohere Analysis,” Cayley’s 
‘‘ Elliptic Functions,” Meyer’s ‘‘ Bestimmte Integrale,” For- 
syth’s ‘‘ Differential Equations,” and Williamson’s ‘‘ Integral 
Calculus.” 

The chapter on Theory of Functions was sketched out and 
in part written by Professor B. O. Peirce, to whom I am 
greatly indebted for numerous valuable suggestions touching 
other portions of the book, and who has kindly allowed me’ 
to have his Short Table of Integrals bound in with this volume. 


W. E. BYERLY. 
CAMBRIDGE, 1888. 


ANALYTICAL TABLE OF CONTENTS. 


CHAPTER I. 
SYMBOLS OF OPERATION. 
Article, 
1. Functional symbols regarded as symbols of operation 


sa Red eh phi gaa eed 


_ 
> 


11. 


12. 
13. 
14, 


15. 
16. 
li. 
18. 


19. 


Compound function; compound operation . 

Commutative or relatively free operations . . 

Distributive or linear operations Bas 

The compounds of distributive ODERALIONE. are ‘alstr ibutive eae 

Symbolic exponents 

The law of indices ; 

The interpretation of a zero ea perien : 

The interpretation of a negative exponent . . ; 

When operations are commutative and distributive, Te vne 
‘ bols which represent them may be combined as if they were 
algebraic quantities 


CHAPTER II. 
IMAGINARIES. 


Usual definition of an imaginary. Imaginaries first forced upon 
our attention in connection with quadratic equations. . 

Treatment of imaginaries purely arbitrary and conventional . 

/—1 defined as a symbol of operation . 

The rules in accordance with which the UAB Se eu — “it is steed. 
V—1 distributive and commutative with symbols of quantity . 

Interpretation of powers of V—1 

Imaginary roots of a quadratic Se eee 

Typical form of an imaginary. Equal imaginaries 

Geometrical representation of an imaginary. feals wh pure 
imaginaries. An interpretation of the operation Nim] 

The sum, the product, and the quotient of two Piers 
a+b V—1 and c+dV-—1, are imaginaries of the typical 
oy NES Se ee ee aa ee aS ar ae erence 


La~| 
a 
ope nw wd © — — 


oo 


aann 


10 


vi INTEGRAL CALCULUS. ts 

Article 

20. Second typical form r(cos¢+ V—l1sing). Modulus and argu- 
ment. Absolute value of an imaginary. Examples ..... 

21. The modulus of the sum of two peat) is never greater 
than the sum of their moduli... i. 2) . -ss.e seu 

22. Modulus and argument of the product of i imaginaries : 

23. Modulus and argument of the quotient of two imaginaries . 

24. Modulus and argument of a power of an imaginary. .... . 

25. Modulus and argument of a root of an imaginary. Example . 


26. Relation between the n nth roots of a real or an imaginary . . 
27. The imaginary roots of land—1. Examples ........ 
28. Conjugate imaginaries. Examples . . .. . so Seen 
29. Transcendental functions of an ae eee variable best defined 
by the aid of series. ... ‘ «ie ot peeee 
30. Convergency of a series contented rey terms: 9 258 ieee 
31. Exponential functions of animaginary. Definition of e? where 
zis imarinary<. 205 i. ae rere © 
32. The law of indices holds tor rene: exponentaae Bxaionie 
33. Logarithmic functions of an imaginary. Definition of log z. 
Log za periodic function. Example. .: 9. 0 geese 
34. Trigonometric functions of an imaginary. Definition of sinz 
and cos.z. “Example <9; 0.0) <4 o 6+ «ta we eee 
35. Sinz anc cosz expressed in enone fore The fundamen- 
tal formulas of Trigonometry hold for imaginaries as well as 
for reals. Examples... . Pir Pent) sgt 
36. Differentiation of Functions of Tinasinaes Variahiee The de- 
rivative of a function of an imaginary is in ae indetermi- 
TSC eet eS a ig nee o Piel 2ectey ° 
37. In aideroutingin Sie may yaaa the Sat ‘Wie a PRN ee 
Example. Two forms of the differential of the independent 
WAYIADIC ice vues emer decyl 
38. Differentiation of a power ore: © |b 0 ne” ea sual) i ane 
$9. Differentiation of.e*.. Example .-. < .) > 2) <ie seen 
40. Differentiation of log:2 5° .°.. °°. 4. tke 
41. Differentiation of sinz and cosz. «She. en 
42. Formulas for direct integration (1., Art. 74) hold when @ is 
Imaginary <6. ws ww wk PR nw ete 
43. Hyperbolic Functions ....... ¢ @ sas ee ee 
44. Examples. Properties of Hyperbone F ination 
45. Differentiation of Hyperbolic Functions ........ 
46. Anti-hyperbolic functions. Examples.......... 
47. Anti-hyperbolic functions expressed as logarithms ...... 
48. Formulas for the direct integration of some irrational forms . 


Page. 


10. 


11 
12 
13 
13 
14 
14 
15 
17 


17 
18 


19 
20 


21 


22 


22 


24 


24 
25 
26 
26 
26 


27 
27 
28 
28 
28 
29 
30 


TABLE OF CONTENTS. 


CHAPTER III. 


GENERAL METHODS OF INTEGRATING. 


Article. 

49. Integral regarded as the inverse of a differential ...... 

50. If fe is any function whatever of x, fx.dx has an integral, and 
but one, except for the presence of an arbitrary constant . 

51. A definite integral contains no arbitrary constant, and is a func- 
tion of the values between which the sum is taken. Exam- 
ete is... See then ee Cao 

52. Definite integral of 2 a AG On Hlous fimetion 

Hon oeormulas for direct integration .-. .0. 1. 606 es 

54. Integration by substitution. Examples - 

55. Integration by parts. Examples. hiiguallausous exaiipies in 
EME NMERCT IR Re Sayre path tr) Meat igh gs 24 lig taeda’ gla he 

CHAPTER IV. 
RATIONAL FRACTIONS. 

56. Integration of a rational algebraic polynomial. Rational frac- 
tions, proper and improper ms 

57. Every proper rational eabtton can be ie ay a sum of sim- 
pler fractions with constant humerators . 

58. Determination of the numerators of the partial thectiddie te 
indirect methods. Examples... . ihe 

59. Direct determination of the numerators of the partial fr yoibts 

60. Illustrative examples . 

61. Illustrative example .... at Os ay 

62. Integration of the partial frac Pie’ 

63. Treatment of imaginary values which may occur in ‘the pave 
rections. Exatiples ...9. . 1. ss swe 

CHAPTER V. 
REDUCTION FORMULAS. 

64. Formulas for raising or ee the exponents in the form 
am™-l(a+ bar)P dx senate ledge Ts 

65. Consideration of special cases. Examples. . 


Vii 


Page. 


32 


32 


40 


40 


42 
43 
45 
46 
48 


49 


52 
54 


vl INTEGRAL CALCULUS. errs 


Article Page. 
20. Second typical form r(cos¢+ V—1sin®@). Modulus and argu- 
ment. Absolute value of an imaginary. Examples ..... 10 
21. The modulus of the sum of two piace. is never greater 
than the sum of theirmoduli ..... 0, «8 2 le 
22: Modulus and argument of the product of imaginaries. .... 12 
23. Modulus and argument of the quotient of two imaginaries .. 13 
24. Modulus and argument of a power of an imaginary. ..... 18 
25. Modulus and argument of a root ofan imaginary. Example . 14 
26. Relation between the n nth roots of a real or an imaginary .. 14 
27. The imaginary roots of land—1. Examples ........ 165 
28. Conjugate imaginaries. Examples. .. .7. . 5 2) (sey 
29. Transcendental functions of an se as variable best defined 
by the aid of series . Sst es oe ale hae yp ae 
30. Convergency of a series éontamies imaginary tart oh ees 
31. Hxponential functions of animaginary. Definition of e where 
zisimaginary ... Mr oe he 
32. The law of indices holds Font imaginary exponeHenee Example 20 
33. Logarithmic functions of an imaginary. Definition of log z. 
Log za periodic function. Example..... Ee xy! 
84. Trigonometric functions of an imaginary. Definition of sinz 
and cos z. -Example so... . wo. 00% sys) se ee 
35. Sinz anc cos z expressed in exponential Porn The fundamen- 
tul formulas of Trigonometry hold for imaginaries as well as 
for reals. Examples... . Sas Siteal ee 
36. Differentiation of Functions of fiaeiire Variables The de- 
rivative of a function of an imaginary is in ae indetermi- 
Nate ey : » | eee ae: 
ore un differentiating. we may ae the PY oy ie a danerean fictie 
Example. Two forms of the differential of the independent 


Verlabie=; <9 3-.5.% rere es 
38. Differentiation of a boner ae Urea ere 
39. Differentiation of.e*.: Example .°. . . . = sia puemseennnnenn 
40. Differentiation oflogz 5... .. 2.5% 6 sn 
41. Differentiation of sinz and cosz. . .°.°. .). 71) a) ee 


42. Formulas for direct integration (1., Art. 74) hold when @ is 
Imaginary ©. 603 ew ww 3 Ue eens. « mol ge en 
43. Hyperbolic Functions, . . 2°. sis % 0 6 pyle eae: 
44. Examples. Properties af Fyperinie Funciioas 0 8 a ogee Uae 
45. Differentiation of Hyperbolic Functions .........%. » 26 
46. Anti-hyperbolic functions. Examples... . oe ee 
47. Anti-hyperbolic functions expressed as jogartthae a a eee 
48. Formulas for the direct integration of some irrational forms . 30 


TABLE OF CONTENTS. Vii 


CHAPTER III. 


GENERAL METHODS OF INTEGRATING. 
Article. Page. 
49. Integral regarded as the inverse ofa differential ....... 82 


50. If fx is any function whatever of x, fx.dx has an integral, and 
but one, except for the presence of an arbitrary constant . . 32 

51. A definite integral contains no arbitrary constant, and is a func- 
tion of the values between which the sum is taken. Exam- 


Pree iol”... Som Jalnes MME HNORGISE, 9: 7b ge has er Seen 
52. Definite integral aes a ie sanaabls: Pantin Ay seltn dd soni’ einen Veber 
Paeeriiaine fOr direct integration (1.0. 2). Woot eee ew ee BE 
54. Integration by substitution. Examples .... P 36 


55. Integration by parts. Examples. Migeellancous: eiari wine in 
SRRMNCS em aid ee hth) AO ome tse alg SOM ee an alte DOE 


CHAPTER IV. 
RATIONAL FRACTIONS. 


56. Integration of a rational algebraic polynomial. Rational frac- 


tions, proper and improper. .. . . « 
57. Every proper rational fraction can be Réduchd ag a sum of sim- 
pler fractions with constant numerators. .. . 40 
58. Determination of the numerators of the partial fractions by 
indirect methods. Examples... . 42 
59. Direct determination of the numerators of the partial pueblos 43 
PRIMER RAW COPE AINIICM? Gere ay 3 oles 6 6 en ee cw te we 46 
61. Illustrative example .... ER ee Oe a as mae 8 
62. Integration of the partial frac Hons: Sia eee 48 
63. Treatment of imaginary values which may occur in the arta 
ET rer RELIG S ital x eile age & fe gw ca lees (et Se 
CHAPTER V. 


REDUCTION FORMULAS. 


64. Formulas for raising or lowering the exponents in the form 
IEICE Umer gee, Chi NAS sn tas pide iy ve. ailcd od) bua bed ve SBS 
65. Consideration of special cases. Examples.......... 54 





Vill INTEGRAL CALCULUS. 
CHAPTER VI. 
IRRATIONAL FORMS. 
Article. : Page. 
66. Integration of the form f(«, Va+bx)dx. Examples ..... 56 
67. Integration of the form f(x, Vet Va+ bs)dx. Examples .. 57 
68. Integration of the form f(x, Va+bu+cu*)dx... 1... oat On 
69. Illustrative example. Examples .....4.:5 «sss <6 «8 59 
70. Integration of the form / (« V4 jae Example «.:. 2. 61 
m 

71. Application of the Reduction Formulas of Chapter V. to irra- 

tional.forms. Examples .. ... «.. . . ‘sss =n 61 


72. 


73. 
74. 
75. 
76. 


ti 
78. 
79. 


80. 


81. 


82. 


Reduction formulas for 


A function rendered irrational through the presence under 
the radical sign of a polynomial of higher degree than the 
second cannot ordinarily be integrated. Elliptic Integrals . 62 


CHAPTER VII. 
TRANSCENDENTAL FUNCTIONS. 


Use of the method of Integration by Parts. Examples .... 63 


Reduction Formulas for sin*x and cos"x. Examples. .... 64 
Integratisn of (sin-'x)*dx. Examples .°.. . . <== 65 
Use of the method of Integration by Substitution ...... 66 
Substitution of e=tan > in integrating trigonometric forms . 67 
Integration of sin™« cos*z.dz. Examples... . Jaynes 68 





CHAPTER VIII. 
DEFINITE INTEGRALS. 


Definition of a definite integ lie as the limit of a sum of 


infinitesimals ...0. 2 6 2 se Ge bes > eee gc 71 
Computation of a definite integr al as es limit of asum. Illus- 
trative examples. Examples . . .... 3. sje 72 


Usual method of obtaining the value of a definite integral. 
Caution concerning multiple-valued functions. Examples . 76 


Article. 


83. 


84. 


85. 


86. 


91. 


92. 


93. 
94. 
95. 
96. 
97. 
98. 


99. 


TABLE OF CONTENTS. 


rp. 4 


Page. 
Consideration of the nature of the value of fe. dx when fx 
becomes infinite for <=a, or x=), or for some value of x 
between a and b. Illustrative examples - 78° 
b 
Test that must be satisfied in order that fi tx .dx may be 
a 
finite and determinate if fx is infinite for some value of x 
between a and b. Illustrative examples. Examples . 80 
ive) «o 
Meaning of fe Jx.dx. Condition that f fx.dx shall be 
finite and determinate URE Ree Ve tik ate Pe ee ek eo Oe 
Maximum-Mininum Theorem. Proof that certain important 
ae 
definite integrals of the form J. Jx.dx are finite and de- 
terminate. Examples . 85 
. Application of reduction aii to ‘definite futon rales 
Examples . 4 : 88 
. Application of the preihion of Wve tuton - Subst lutte os 
definite integrals. Illustrative examples. Example. 90 
. Differentiation of a definite integral. Examples 93 
. Many ingenious methods of finding the values of definite infes 
grals are valid only in case the es ie is finite and de- 
terminate . Sap ee 96 
Integration uF aeveroninenit fh in series. Examples . 96 
Values ak log sine.de, (~ eae, { ae obtained 
0 
by ingenious devices. Examples 98 
Differentiation or integration with respect tes a Eaantiee which 
is independent of «. Examples. ; 23004 
Additional illustrative examples. Examples . 102 
Introduction of imaginary constants. . 104 
The Gamma Function CORR yas BO el. deta 175 
Table giving logr(n) from n= 1 e n=2. Definite integrals 
expressed as Gamma Functions . rad Nig’ 
The Beta Function. Formula connecting phe Beta Panction 
with the Gamma Function. Value of r(3) . 109 
More definite integrals expressed as Gamma Functions. 
Mxampies..°. +> we ltG 


Article. 
100. 
101. 
102. 
103. 
104. 
105. 
106. 
LOT « 
108. 
109. 
110. 
Lit. 
112. 
113. 
114. 
115. 
LG. 
117. 


INTEGRAL CALCULUS. 


CHAPTER IX. 


LENGTHS OF CURVES. 


Page. 


Formulas toe Sint and cosr in terms of the length of the arc 113 


The equation of the Catenary obtained. Example 
The equation of the 7'ractriz. Examples. 
Length of an arc in rectangular coordinates . 
Length of the arc of the Cycloid. Example. 
Another method of rectifying the Cycloid . 
Rectification of the Hpicycloid. Examples 

Arc of the Hilipse. Auxiliary angle. Example 
Length of an are. Polar coordinates 

Equation of the Logarithmic Spiral 


Rectification of the Logarithmic Spiral. Examples . 


Rectification of the Cardioide , 
Involutes. Illustrative example. Example 
The involute of the Cycloid. Example . 
Intrinsic equation of a curve. Example 
Intrinsic equation of the Epicycloid. Example 
Intrinsic equation of the Logarithmic Spiral. 


in rectangular coordinates. Examples . 


. Intzivsic eauation of an evolute 

. Illustrative examples. Examples 

. The evolute of an Epicycloid. Example 
. The intrinsic equation of an involute. Illustrative bcheatae 
. Limiting form approached by an involute of an involute 

. Method of obtaining the equation in rectangular coordinates 
from the intrinsic equation. ITllustrative example . 
. Rectification of Curves in Space. Examples. 


CHAPTER X. 


AREAS. 


. Areas expressed as definite integrals, rectangular codrdinates. 


Examples . 


. Areas expressed as definite ‘atari pains coérdinaees 
7. Area between the catenary and the axis : 
. Area between the tractrix and the axis. Example 


~ 118 
pe 8B 3 
ELT 
. 118 
A190 
- 119 
. 120 
ne | 
. 121 
. 122 
. 122 
. 123 
. 125 
. 126 
Amr § 
. 128 
Method of obtaining the intrinsic equation from the canton 

. 128 
. 180 
. 130 
» 131 


132 


. 133 


. 134 
. 185 


oer 
. 139 
. 139 
. 139 


TABLE OF CONTENTS. ~ xi 


Article. th Page. 


129. Area between a curve and its asymptote. Examples .. . 140 
130. Area of circle obtained ee the aid of an oa angle. Ex- 

amples. so . =. beaten rec oo eee eee & | 
131. Area between two curves nea coor. ). Examples... . 142 
132. Areas in Polar Codrdinates. Examples .. . . 148 
133. Problems in areas can often be simplified by franaroriiation 

BprOuruiniven. MAA Pes os se a ae AEE 
134. Area between a curve and its evolute. Examples. .. . . 146 
1385. Holditch’s Theorem. Examples .. . OAL Pe Sayeeda i vi 
136. Areas by a double integration (rect. coor. ie en ghia eke eae eu 
da7- iiinstrative examples. Examples. *. ....... .«. «150 
138. Areas by a double integration (polar codr.). Example . . . 151 
e 


CHAPTER XI. 


AREAS OF SURFACES. 


139. Area of a surface of revolution (rect. coor.). Example . . . 158 
140. Illustrative examples. Examples. .. . . 154 
141. Area of a surface of revolution by erase rorramilan of nite 
nates. Example .... da tie ae neces LD 
142. Area of a surface of revolution Hisar coor. * seaples ares BY 
143. Area of acylindrical surface. Examples ....... . 157% 
144. Area of any surface by a double integration . .. . . . . 160 
145. Illustrative example. Examples... . 163 
146. Illustrative example requiring Pcie tormation ‘a. NAPS éoardt: 
PTO ICHUINOH Sc oto oii e h es a. «ea oe, © LBS 
. CHAPTER XII. 
VOLUMES. 
147. Volume by a single integration. Example ...... . 168 
ee vomme Of a conocid.. Examples . 9:0. 2. ee we. 169 
149. Volume of an ellipsoid. Examples .. . enue ben aa 
150. Volume of a solid of revolution. Single inteevatton: Exam- 
CT i ae ee eee Gh Sai ay lee tes Ce ey 
151. Volume of a solid of nei nton Double integration. Exam- 
Dies ses a sek eS oheehiet es bapeaeremargy Wp 


152. Volume of a solid of fea listéots Polar formula. Example . 174 

153. Volume of any solid. Triple integration. Rectangular coor- 
dinates. Examples .. . . 175 

154. Volume of any solid. Triple ra aeee tical: Peat puahainates 
OAT WEES ON a AUD i et a Lena ee er ee re Ab 


Xi 


Article. 
155. 


156. 
157. 
158. 
159. 
160. 


161. 


162. 
163. 


164. 
165. 
166. 
167. 


168. 


169. 
170. 
1 6s 


172. 


INTEGRAL CALCULUS. 


CHAPTER XIII. 


CENTRES OF GRAVITY. 


Page. 
Centre of Gravity defined. . . . . :. See ee 
General formulas for the coordinates of one Centre of Gravity 
of any mass. Example. .. . ee 
Centre of Gravity of a homogeneous ode » ad Sey 
Centre of Gravity of a plane area. Examples . ... . . 182 
Centre of Gravity of a homogeneous solid of revolution. Ex- 
aminles sx ss oe eae ee led 
Centre of Gravity ne an are ne of a surface of revolution. Ex- 
amples... eee Sg oe el rr 
Properties of Guldin. Rueantples ek 8) wd aa ees 
CHAPTER XIV. 
LINE, SURFACE, AND SPACE INTEGRALS. 
Point function. Continuity of a point function. . . . 190 
Line-integral, surface- Beer and space-integral of a pone 
LMC OU Uy tas ; . 190 


Value of a line, sniteces or space iuteora indepentaa of the 
position in each element of the point at which the value of 
the function is taken. .. . a 
Value of a line, surface, or space inte - indepéncent of the 
manner in which the line, surface, or space is broken up 


into infinitesimal elements. . . . 191 
Geometrical representation of a line- cera alone a alan 

curve, and a surface-integral over a plane surface. . . . 193 
Moments of inertia. Examples. .. . . 193 


Relation between a surface-integral over a ae sur face aiifl 
a line-integral along the curve bounding the surface. Ex- 
RINDIGE 2 se : Soe ae ee ne 
Illustrative acne ReaolEe ms hee « ae ee EG 
Another form of the relation established in Art. 168. Example 199 
Relation between a space-integral taken throughout a given 
space and a surface-integral over the surface bounding the 
space. Example... . . %. .. ss. \ae 5) Qe 
Illustrative example, Example . . . «...)) (2 een 


Article. 
173. 
174. 


175. 


176. 
177. 
178. 
179. 


180. 


181. 


182. 
183. 


184. 


TABLE OF CONTENTS. Xi. 


CHAPTER XV. 


MEAN VALUE AND PROBABILITY. 
Page. 
References .. . ; Sael gat took BOL 
Mean value of a paneeope ttf varying annitiey The mean ne 
tance of all the points of the circumferences of a circle from 
a fixed point on the circumference. The mean distance of 
points on the surface of a circle from a fixed point on the 
circumference. The mean distance of points on the surface 
of a square from a corner of the square. The mean distance 


between two points within a givencircle ... . . 202 
Problems in the application of the Integral Calculus “ha pr thas 
bilities. Random straight lines. Examples ..... . 204 


CHAPTER XVI. 
ELLIPTIC INTEGRALS. 


Motion of a simple pendulum. Vibration. Complete revolution 211 


The length of anarc of an Ellipse. . . . 218 
Algebraic forms of the Filiptic Integrals of ie ne Beco 
and third class. Modulus. Parameter. ... . y « 2k3 
Trigonometric forms of the Elliptic Integrals. Amplitude. 
Delta. Complementary Modulus .. . . 214 


Landen’s Transformation. Reduction eonnula ne Whitets we 
can increase the modulus and diminish the amplitude of an 
Elliptic Integral of the first class. A method of free 
‘of Oy 2 ee ae . 215 

Reduction formula for diminishing the podatta and: increas- 
ing the amplitude of an Elliptic Integral of the first class.’ 


A second method of computing F(k,~). . . . .. . . 218 
Actual computation of F( ed 4 and ers v2 2) rh RS 3 


Landen’s Transformation. Reduction formula by which we 
can increase the modulus and diminish the amplitude of an 
Elliptic Integral of the second class. A method of com- 
puting E(k,9¢).. . . 222 

A reduction formula for atriniantne the meninges bane in- 
creasing the amplitude of an Elliptic Integral of the second 
class. A second method of computing H(k,o). .. . . 225 


X1V 


Article. 


185. 


186. 


187. 


188. 


189. 
190. 
191. 
192. 


193. 
194. 


195. 


196. 
197. 


198. 


199. 


200. 
201. 


202. 
203. 


INTEGRAL CALCULUS. 


Page. 


Actual computation of £(, % and her v2 zy 5 ean 


An Elliptic Integral of the first or second class, whose ampli- 


tude is greater than ~ can be made to depend upon one 


whose amplitude is less than 3 and upon the correspond- 


ing complete Elliptic Integral . .. .- . 231 
Three-place table of Elliptic Integrals of the first chaga and 

the second class .. . sleet Reva bane ~ 9) iO eee 
Addition Formulas. Wanetions defined by the aid of definite 

integrals. log«, sin“, tan-1e, F(x, a E i si Addition 

formula forlogx . . . : : . « 235 
Addition formulas for sin™ “3 ae tan” oe MEM ee 
Addition formula for F(%,x7) .. . ge mae 
Analogy between log”! u, sin uw, tan u, ana F- 1 (it, ase sh ee LeEL 
The Elliptic Functions, snu, cnu, and dnu. Their analogy 

with Trigonometric Functions. Formulas CT 

Elliptic Functions of a single quantity . . . . . . 242 
Formulas for Elliptic Functions of (uw+v) and (ee . . 244 
Formulas for Elliptic Functions of 2u...... .. . . 246 


Formulas for Elliptic Functions of | MPa ey 


Periodicity of the Elliptic Functions. Real period,4A. . . 247 

Elliptic Functions of a pure imaginary. Jacobi’s Transforma- 
tion. Elliptic Functions have an imaginary period, 4 K'V—1. 
Table of values of Elliptic Functions having the modulus 


VQ a wi eh ek re 

2 
The Elliptic Integral of the second class expressed in terms 

of Elliptic Functions. Addition formula for ee Inte- 

grals of the second class .. . oS ee eae 
Application to the rectification of fe L ont Bxcanpite 

Bisection of the arc of a quadrant of the Lemniscate. . . 254 
Rectification of the Ellipse. Examples. .. . oon. 
Use of the Addition Formula in dealing with Elliptic arcs. 

Fagnani’s Point. Examples . . .° . 4) 3. 2 ese 
Rectification of the Hyperbola. Examples ...... . 260 
The simple pendulum. Examples . .°-.° 5° 2) sep 


Article. 


204. 
205. 


206. 
207. 
208. 


209. 
210. 


att; 
212: 


213. 


214, 


215. 


216. 


217. 


218. 


219. 


220. 


221. 


TABLE OF CONTENTS. XV 


CHAPTER XVII. 


INTRODUCTION TO THE THEORY OF FUNCTIONS. 

Page. 

Single-valued functions. Multiple-valued functions . . . . 267 
Importance of the graphical representation of imaginaries. 

Complex quantity . . . met gr 7) 

When a complex variable is ald a Ape eR, hoch ae ee iy pe 

A continuous function of a complex variable. Critical values 268 
Criterion that a function shall have a determinate derivative. 

Monogenic functions. . . ’ oe ee, 

Any function involving z asa whniet is a Monogenic finetion 272 


Conjugate functions. Their use as solutions of Laplace’s 

EB MEM SOUND on * le hey tis ge Wet ales ears) ce OED 
Preservation of angles . . . . 274 
If two paths traced by the eink precsinay the annie 


have a common beginning and a common end, and do not 
enclose a critical point, the corresponding paths traced by 
the point representing the function and having a common 


beginning willhaveacommonend ... . . 276 
Examples where the paths traced by the point Scireacc nies 
the variable enclose a critical point . .. . . 278 


Critical points at which the derivative of the fanstion is zero 
or infinite are to be avoided. Branch points. Holomorphic 
functions . .. . 279 
Definite integral of a eeaneiion of ¢ a a vielanle deinen: 
Such an integral is generally indeterminate, and depends 
upon the path by which the point representing the variable 
passes from the lower limit to the upper limit of the integral 281 
If the function is holomorphic, the definite integral is in gen- 


eral determinate... . eae 
The integral around a iosed Kato, useage a ee at 

mecmrerte: cnnction IS infinite: ..56.. «ts («wie p84 
Illustrative examples .. . . 285 


Convergency of the series SE Guied iecsratie fis tie 
of a convergent series where the separate terms are holo- 


morphic functions. .. . . 287 
Proof of Taylor’s and Maciairin? Ss eyes roe for finctions of 
complex variables. Circle of convergence . . . . 288 


Investigation of the convergency of various series hich are 
obtained by Taylor’s and Maclaurin’s Theorems. Examples 291 


Xvl INTEGRAL CALCULUS. 


CHAPTER XVIII. 


KEY TO THE SOLUTION OF DIFFERENTIAL EQUATIONS. 
Article. 


Page. 

222. Description of Key ... . Pa Pas Sle 4) 
223. Definition of the terms differential OquGHOn hoe ews 
linear, general solution. or complete primitive, singular solu- 

tion, exact differential equation . ... ~ a2) 05 seen 

224. Examples illustrating the use of the Key . . . on Le papeae 

225. Simplification of differential equations by change of variable . 308 

FO 0 es eat PPP 


Reiners Capes MPP 


INTEGRAL CALCULUS. 


CHAPTER I. 
SYMBOLS OF OPERATION. 


1. It is often convenient to regard a functional symbol as 
indicating an operation to be performed upon the expression 
which is written after the symbol. From this point of view the 
symbol is called a symbol of operation, and the expression writ- 
ten after the symbol is called the subject of the operation. 

Thus the symbol D, in D,(a’y) indicates that the operation of 
differentiating with respect to x is to be performed upon the 
subject (xy). 


2. If the result of one operation is taken as the subject of a 
second, there is formed what is called a compound function. 

Thus. logsina is a compound function, and we may speak of 
the taking of the logsin as a compound operation. 


3. When two operations are so related that the compound 
operation, in which the result of performing the first on any 
subject is taken as the subject of the second, leads to the same 
result as the compound operation, in which the result of per- 
forming the second on the same subject is taken as the subject 
of the first, the two operations are commutative or relatively free. 

Or to formulate ; if 

ie Es 


the operations indicated by f and F’ are commutative. 


2 INTEGRAL CALCULUS. [ART. 4. 


For example; the operations of partial differentiation with 
respect to two independent variables x and y are commutative, 
for we know that . 


D,D,u= D,D,%. (I. Art, toe 
The operations of taking the sine and of taking the logarithm 
are not commutative, for logsinu is not equal to sinlogw. 


4, If f(utv)=fuxt fv 


where uw and v are any subjects, the operation fis distributive or 
linear. 

The operation indicated by d and the operation indicated by 
D, are distributive, for we know that 


d(u+tv)=du +t dv, 
and that VOGT a ah) ents BN + es 


The operation sin is not distributive, for sin(w+v) is not 
equal to sinu + sinv. 


5. The compounds of distributive operations are distributive. 
Let f and F indicate distributive operations, then fF' will be 
distributive ; for 
Huo) a ep, 


therefore /fF(utv)=f(Fut Fv)=fFut fF. 


6. The repetition of any operation is indicated by writing an 
exponent, equal to the number of times the operation is per- 
formed, after the symbol of the operation. 

Thus log’? means loglogloga; d’u means dddu. 

In the single case of the trigonometric functions a different 
use of the exponent is sanctioned by custom, and sin?w means 
(sinw)? and not sin sinw. 


7. Ifm and n are whole numbers it is easily proved that 


A eh te Le 


Cuap. I.] SYMBOLS OF OPERATION. 3 


This formula is assumed for all values of m and n, and nega- 
tive and fractional exponents are interpreted by its aid. It is 
called the law of indices. 


8. To find what interpretation must be given to a zero ex- 


eu? let m= 0 in the formula of Art. 7. 


0 Bag) 9 Baie haa) Us 
or, denoting f”u by v, FOU: 


That is; a symbol of operation with the exponent zero has no 
effect on the subject, and may be regarded as multiplying it by 
unity. 


9. To interpret a negative exponent, let 


m=-—n in the formula of Art. 7. 
mie re aap. Uh — th 
If we call Pras a Ue CAN yo Wem es 
thie i seen | 
we get anf th Sarl 


and the exponent —1 indicates what we have called the anti- 
function of fu. (1. Art. 72.) 

The exponent —1 is used in this sense even with trigonometric 
functions. 


10. When two operations are commutative and distributive, 
the symbols which represent them may be combined precisely as 
if they were algebraic quantities. 

For they obey the laws, 


a(m+n)=am+an, 
am = ma, 


on which all the operations of arithmetic and algebra are founded. 


4 INTEGRAL CALCULUS. (Arr. 10. 


For example ; if the operation (D,+ D,) is to be performed 
n times in succession on a subject wu, we can expand (D,+ D,)” 
precisely as if it were a binominal, and then perform on w the 
operations indicated by the expanded expression. 


(D, + D,)?u= (D2 +3 DZD, +3 D,D/ + DZ)u 
= Diu+3 D?D,u+3D,D7u+ Du. 


Cuap. II.] IMAGINARIES. 5 


CHAPTER II. 
IMAGINARIES. 


11. An imaginary is usually defined in algebra. as the indi- 
cated even root of a negative quantity, and although it is clear 
that there can be no quantity that raised to an even power will 
be negative, the assumption is made that an imaginary can be 
treated like any algebraic quantity. 

Imaginaries are first forced upon our notice in connection 
with the subject of quadratic equations. Considering the typical 
quadratic pe age Hes 0: 
we find that it has two roots, and that these roots possess cer- 
tain important properties. For example; their sum is —a@ and 
their product is 0. We are led to the conclusion that every 
quadratic has two roots whose sum and whose product are 
simply related to the coefficients of the equation. 

On trial, however, we find that there are quadratics having 
but one root, and quadratics having no root. 

For example ; if we solve the equation 


we find that the only value of # which will satisfy it is unity; 
and if we attempt to solve 


v—2ae+2=—0, 


we find that there is no value of x which will satisfy the equation. 
As these results are apparently inconsistent with the conclu- 
sion to which we were led on solving the general equation, we 
naturally endeavor to reconcile them with it. 
The difficulty in the case of the equation which has but one 


6 INTEGRAL CALCULUS. (ART. 12. 


root is easily overcome by regarding it as having two equal roots. 
Thus we can say that each of the two roots of the equation 


g—-%+t1l=O0 


is equal to 1; and there is a decided advantage in looking at the 
question from this point of view, for the roots of this equation 
will possess the same properties as those of a quadratic having 
unequal roots. The sum of the roots 1 and 1 is minus the co- 
efficient of x in the equation, and their product is the constant 
term. 

To overcome the difficulty presented by the equation which 
has no root we are driven to the conception of imaginaries. 


12. An imaginary is not a quantity, and the treatment of 
imaginaries is purely arbitrary and conventional. We begin by 
laying down a few arbitrary rules for our imaginary expressions 
to obey, which must not involve any contradiction; and we 
must perform all our operations upon imaginaries, and must 
interpret all our results by the aid of these rules. 

Since imaginaries occur as roots of equations, they bear a close 
analogy with ordinary algebraic quantities, and they have to be 
subjected to the same operations as ordinary quantities; there- 
fore our rules ought to be so chosen that the results may be 
comparable with the results obtained when we are dealing with 
real quantities. 


13. By adopting the convention that 
V—@=av—1, 


where «@ is supposed to be real, we can reduce all our imaginary 
algebraic expressions to forms where -V —1 is the only peculiar 
symbol. This symbol V —1 we shall define and use as the sym- 
bol of some operation, at present unknown, the repetition of which 
has the effect of changing the sign of the subject of the operation. 
Thus in a V —1 the symbol V—1 indicates that an operation 
is performed upon @ which, if repeated, will change the sign 


of a. That is, ie 
aN 1)? =a 


Cuap. II.] IMAGINARIES. 7 


From this point of view it would be more natural to write the 
symbol before instead of after the subject on which it operates, 
(V—1)a instead of aV—1, and this is sometimes done; but 
as the usage of mathematicians is overwhelmingly in favor of the 
second form, we shall employ it, merely as a matter of con- 
venience, and remembering that a is the subject and the V —1 
the symbol of operation. 


14. The rules in accordance with which we shall use our new 
symbol are, first, 


aV —1+bV —1=(a+b)V—1. [1] 


In other words, the operation indicated by V —1 is to be dis- 
tributive (Art. 4) ; and second, 

ON ol (V1) a, [2] 
or our symbol is to be commutative with the symbols of quantity 
(Art. 3). 

These two conventions will enable us to use our symbol in 
algebraic operations precisely as if it were a quantity (Art. 10). 

When no coefficient is written before V —1 the coefficient 1 
will be understood, or unity will be regarded as the subject of 
the operation. 


15. Let us see what interpretation we can get for powers of 
\/—1; that is, for repetitions of the operation indicated by the 
symbol. 


(v—1)'=1 (Art. 8), 
(V—1)'= V—1, 
(v—1)?=—-1, by definition (Art. 13), 


(¥—1)?= (V—1)?°V=1=—V-=1, by definition, 
An ain eo te 

(V=>1)'=1V—1 =V-1, 

eo Ea hy ed, 


and so on, the values V erent ama coon e.' f) occurring in 
cycles of four. 


8 INTEGRAL CALCULUS. [ArT. 16. 


16. The definition we have given for the square root of a 
negative quantity, and the rules we have adopted concerning its 
use, enable us to remove entirely the difficulty felt in dealing 
with a quadratic which does not have real roots. ‘Take the 
equation 


xe? —2¢+5=0. (1) 
Solving by the usual method, we get 
Seah Wein af seed es 
> /—4=2V—1, by Art. 13 [1]; 
hence es 1 Ba or be Bods 


On substituting these results in turn in the equation (1), per- 
forming the operations by the aid of our conventions (Art. 14 
[1] and [2]), and interpreting (V —1)? by Art. 15, we find that 
they both satisfy the equation, and that they can therefore be 
regarded as entirely analogous to real roots. We find, too, that 
their sum is 2 and that their product is 5, and consequently that 
they bear the same relations to the coefficients of the equation as 
real roots. 


17. An imaginary root of a quadratic can always be reduced 
to the form a + bV —1 where a and 0 are real, and this is taken 
as the general type of an imaginary; and part of our work will 
be to show that when we subject imaginaries to the ordinary 
functional operations, all our results are reducible to this typical 
form. 


If two imaginaries a+bV—1 and e+dV—1i1 are equal, 
a must be equal to c, and 6 must be equal to d. 

For we have atobV—-l=c+dvV—1. 

Therefore a—c =(d—b)V—1, 


or a real is equal to an imaginary, unless a—c=0=d—b. 
Since obviously a real and an imaginary cannot be equal, it 
follows that a=c and b=d. 


Cuap. II.] IMAGINARIES. 9 


18. We have defined V —1 as the symbol of an operation 
whose repetition changes the sign of the subject. 

Several different interpretations of this operation have been 
suggested, and the following one, in which every imaginary is 
graphically represented by the position of a point in a plane, is 
commonly adopted, and is found exceedingly useful in suggest- 
ing and interpreting relations between different imaginaries and 
between imaginaries and reals. 

In the Calculus of Imaginaries, a+ b-/ —1 is taken as the 
general symbol of quantity. If b is equal to zero,a+bV—1 
reduces to a, and is real; if a is equal to zero, a+b~V —1 re- 
duces to b-V —1, and is called a pure imaginary. 

a+b~—1 is represented by the position of a point referred 
to a pair of rectangular axes, as in analytic geometry, a being 
taken as the abscissa of the 
point and 6 as its ordinate. 
Thus in the figure the position 
of the point P represents the ye; 
imaginary a+b ee 


Y 


If 6=0, and our quantity is 
Pencil lib. on tie: axig)oL, oe 
X, which on that account is 
called the avis of reals ; if a=0, 
and we have a pure imaginary, 
P will lie on the axis of =X, nd ¢ 
which is called the axis of pure imaginaries. 

It follows from Art. 17 that if two imaginaries are equal, the 
points representing them will coincide. 

Since a and aV—l1 are represented by points equally distant 
from the origin, and lying on the avis of reals and the axis of 
pure imaginaries respectively, we may regard the operation 
indicated by V —1 as causing the point representing the subject 
of the operation to rotate about the origin through an angle of 
90°. A repetition of the operation ought to cause the point to 
rotate 90° further, and it does; for 

a(V —1)?=—a, 


and is represented by a point at the same distance from the 





10 INTEGRAL CALCULUS. (Arr. 19. 


origin as a, and lying on the opposite side of the origin; again 
repeat the operation, 


a(n —1)8=—a Vien, 
and the point has rotated 90° further; repeat again, 
a(V —1)*=a, 


and the point has rotated through 360°. We see, then, that if 
the subject is a real or a pure imaginary the effect of performing 
on it the operation indicated by V —1 is to rotate it about the 
origin through the angle 90°. We shall see later that even when 
the subject is neither a real nor a pure imaginary, the effect of 
operating on it with V —1 is still to produce the rotation just 
described. 


19. The sum, the product, and the quotient of any two imagi- 
naries,a+bV—1 andc+dv—l, are imaginaries of the typi- 
cal form. 


atoV—1l+e+dV—-1 =a+c+(b+d)vV—1. [1] 
(a +bV—1) (ec t+dV—1) =ac—bd+ (be+ad)V—1.  f2) 


at+bV—1 _ (a+bV—1) (e—dV—1) _ act+bd+ (be—ad)V—1 
etdV—1 (c+dV—1) (ec—dV—1) o+@ 


_ ac + bd be — ad 
4g! +e 








ele [3] 
All these results are of the form 4+BvV—1. 


20. The graphical representation we have suggested for 
imaginaries suggests a second typical form for an imaginary. 
Given the imaginary «+y7V—1, let the polar codrdinates of 
the point P which represents «+ yV—1 be r and ¢. 

r is called the modulus and ¢ the argument of the imaginary. 


— iS 


Cnap. II.] IMAGINARIES, TA 


The figure enables us to establish very 
simple relations between x, y, 7, and ¢. 





: y 
om a 6:5 ie Ui 
r= Vet yf, 
o= tan—! = Bes 
2-+yV—1=rcos¢+ (V—1)rsing 
= r(cos¢+V—1.sing), [3] 


where the imaginary is expressed in terms of its modulus and 
argument. 

The value of 7 given by our formulas [2] is ambiguous in 
sign; and @ may have any one of an infinite number of values 
differing by multiples of z. In practice we always take the 
positive value of 7, and a value of ¢ which will bring the point 
in question into the right quadrant. In the case of any given 
imaginary then, 7 can have but one value, while ¢ may have any 
one of an infinite number of values differing by multiples of 27. 

The modulus r is sometimes called the absolute value of the 


imaginary. 
EXAMPLES. 


(1) Find the modulus and argument of 1; of Men lois Ans 
or — 2 —1 : of 3+-3V—1; of 2+4 V—1; and express each of 
these quantities in the form r(cos#/ +V—1.sin¢). 


(2) Show that every positive real has the argument zero; 
every negative real the argument 7; every positive pure imagi- 
4 T . . . 
nary the argument 5? and every negative pure imaginary the 


argument par 
2 


21. If we add two imaginaries, the modulus of the sum is 
never greater than the swm of the moduli of the given imagi- 
naries. 


12 INTEGRAL CALCULUS. [ART. 22. 


The sum of a+bV—1 and c+dV—1 isatc+ (04-d) Voeae 
The modulus of this sum is V (a+c)?+(6+d)*; the sum of 
the moduli of a+bdV—1 and c+dV—1lisV?4+0?4+Ve4+ @. 
We wish to show that 





VG FOTO! <VEFFLVETE; 
the sign -< meaning ‘‘ equal to or less than.” 
Now V(ate?+(o+d)? <VeLe+Vera, 
if (a+c)?+(b+d)? <@?+0?+2V(e7+0’) (P+ @)4+°4+ 0, 
that is, if actbd -<VeEF+0?C+0?+0 a; 
or, squaring, if 
Veet 2abed +0? <i +e? + he + bd: 
or, if . 0 -< (ad — be)’. 


This last result is necessarily true, as no real can have a 
square less than zero; hence our proposition is established. 


22. The modulus of the product of two imaginaries is the 
product of the moduli of the given imaginaries, and the argument 
of the product is the sum of the arguments of the imaginaries. 

Let us multiply 


1, (COS dy + Kee BIN di) by 1(cosdg+ /—1.sin do) 3 
we get 
71 1'2[ COS , COS Ho— sin d Sin f.+ V — 1(sin 4, cos do+ Cos ¢; sin $2) |, 
COS 1 COS hy — SiN ; Sin dy = COS(, + dy), 
sin 4) COS dz + COS f) Sin $2 = Sin(d; + do) 
by Trigonometry ; hence 
1, (cos d; +V —1. sin d,) 72 (cos Pat \/ —1. sin dy) 
= 1,72 [cos(¢1 + 2) + V—1. sin(¢, + ¢2)], 


Cuap. II.] IMAGINARIES. 13 


and our result is in the typical form, 7,7, being the modulus and 
$, + ¢» the argument of the product. 

If each factor has the modulus unity, this theorem enables us 
to construct very easily the product of the imaginaries ; it also 
enables us to show that the interpretation of the operation V —1, 
suggested in Art. 18, is perfectly general. 

Let us operate on any imaginary subject, 


r(cosd + V—1.sin¢), with V—1, 
that is, with 1 (cos qe Ape sin 5): 


The modulus r will be unchanged, the argument ¢ will be in- 
creased by t and the effect will be to cause the point repre- 


senting the given imaginary to rotate about the origin through 
an angle of 90°. 


23. Since division is the inverse of multiplication, 


1,(cos, + V —1. sing,) + 72(cos¢, + V—1. sind) 


will be equal to 


: [cos (¢1 — ¢2) + V1. sin(¢; — ¢2) }, 


since if we multiply this by 7,(cos¢. + \/—1. sin ds), according 
to the method established in Art. 22, we must get 


7,(cos¢, + V—1.sind)). 


To divide one imaginary by another, we have then to take the 
quotient obtained by dividing the modulus of the first by the 
modulus of the second as our required modulus, and the argu- 
ment of the first minus the argument of the second as our new 
argument. 


24. If we are dealing with the product of n equal factors, or, 
in other words, if we are raising r(cos¢+ V—1.sindg) to the 


14 INTEGRAL CALCULUS. [ARr. 25. 


nth power, n being a positive whole number, we shall have, by 
Art. 22, 


[r(cosd + V—1. sing) ]"=7"(cosnd + V—1.sinnd¢). [1] 


If 7 is unity, we have merely to multiply the argument by n, 
without changing the modulus; so that in this case increasing 
the exponent by unity amounts to rotating the point represent- 
ing the imaginary through an angle equal to ¢ without changing 
its distance from the origin. | 


25. Since extracting a root is the inverse of raising to a 
power, 


'/[r(cosé + V—1.sind) ] = Vr(cos® +V=I. sin) [1] 
for, by Art. 24, 


n 


| a7( cos + Ajo dS Sin a) = r(cos¢ + V—1.sin p). 


EXAMPLE. 


Show that Art. 24 [1] holds even when n is negative or 
fractional. 


26. As the modulus of every quantity, positive, negative, 
real, or imaginary, is positive, it is always possible: to find the 
modulus of any required root; and as this modulus must be real 
and positive, it can never, in any given example, have more than 
one value. We know from algebra, however, that every equa- 
tion of the nth degree containing one unknown has n roots, and 
that consequently every number must have n nth roots. Our 
formula, Art. 25 [1], appears to give us but one mth root for 
any given quantity. It must then be incomplete. 

We have seen (Art. 20) that while the modulus of a given 
imaginary has but one value, its argument is indeterminate and 
may have any one of an infinite number of values which differ by 
multiples of 27. If ¢) is one of these values, the full form of 


_ Cuap. II.) IMAGINARIES. 15 
the imaginary is not r(cos ¢) + V—1.sindy), as we have hitherto 
written it, but is 


7 [cos(¢o +.2 mr) + seme hp sin{dy + 2 mz) ], 


where m is zero or any whole number positive or negative. 
Since angles differing by multiples of 27 have the same trigo- 
nometric functions, it is easily seen that the introduction of the 
term 2m into the argument of an imaginary will not modify 
any of our results except that of Art. 25, which becomes 


Vr [cos (dy -+ 2m) + -V—I. sin(¢y) + 2mrz) ] 
= | cos (S4 m=") + s/—1. sin (B+ m=) Meat k | 





Giving m the values 0,.1,°2, 3 ..... »>r—1,n, n+1, success- 
ively, we get 
do do , 27 do , 527. po 20 Po 2a 
eae on eu nat Ba ge hk Wag tiaras 
2 
SONA AN Sa) aeRO 
1 11) nr 


as arguments of our nth root. 

Of these values the first n, that is, all except the last two, 
correspond to different points, and therefore to different roots ; 
the next to the last gives the same point as the first, and the 
last the same point as the second, and it is easily seen that if we 
go on increasing m we shall get no new points. The same thing 
is true of negative values of m. 

Hence we see that every quantity, real or imaginary, has n 
distinct uth roots, all having the same modulus, but with argu- 


ments differing by multiples of oe 
n 


27. Any positive real differs from unity only in its modulus, 
and any negative real differs from —1 only in its modulus. All 
. the nth roots of any number or of its negative may be obtained 


16 INTEGRAL CALCULUS. [ART. 27. 


by multiplying the mth roots of 1 or of —1 by the real positive 
nth root of the number. 

Let us consider some of the roots of 1 and of —1; for ex- 
ample, the cube roots of 1 and of —1. The modulus of 1 
is 1, and its argument is 0. The modulus a each i the cube 


roots of 1 is 1, and their arguments are 0, =, and & a: ; that is, 
0°, 120°, and 240°. The roots in question, then, are repre- 


sented by the points P,, Ps, P3, in the figure. Their values are 


1(cos0 + V—1. sin0), 


s 1 (cos 120° + V—1. sin 120°), 
p, and 1(cos240° + V—1. sin 240°), 
or 1, -$+2V—1, —f ee 
Ps The modulus of —1 is 1, and its 
argument is z. The modulus of the 
: . 7 oT Pde 
Wes roots of —1 is 1, and their arguments are —, —-++ — 


ert 
a4, that is, 60°, 180°, 300°. The roots in question, then, 
_ are represented by the points P,, P,, 


P;, in the figure. Their values are 


P, EXAMPLES. 


(1) What are the square roots of 
P, land —1? the 4th roots? the 5th 
roots ? the 6th roots ? 


(2) Find the cube roots of —8; the 5th roots of 32. 


(3) Show that an imaginary can have no real nth root; that 
a positive real has two real nth roots if n is even, one if n is 


odd; that a negative real has one real nth root if n is odd, none 
if n is even. 


Cuap. IL. ] IMAGINARIES. \ 17 


28. Imaginaries having equal moduli, and arguments differing 
only in sign, are called conjugate imaginaries. 

r(cosg@ + V—l.sing), and rfcos(—¢) +V—1. sin(—¢) ], 
or 7(cos ¢ — V—1.sind) are conjugate. 

They can be written «+yV—1 and «—yV—1, and we see 
that the points corresponding to them have the same abscissa, 
and ordinates which are equal with opposite signs. 


EXAMPLES. 


(1) Prove that conjugate imaginaries have a real sum and a 
real product. 


(2) Prove, by considering in detail the substitution of 
a+bV—1 and a—bV—1 in turn for 2 in any algebraic poly- 
nomial in # with real coefficients, that if any algebraic equation 
with real coefficients has an imaginary root the conjugate of that 
root is also a root of the equation. 


(8) Prove that if in any fraction where the numerator and 
denominator are rational algebraic polynomials in a, we substi- 
tute a+bV—1 and a—bV—1 in turn for 2, the results are 
conjugate. 


Transcendental Functions of Imaginaries. 


29. We have adopted a definition of an imaginary and laid 
down rules to govern its use, that enable us to deal with it, in 
all expressions involving only algebraic operations, precisely as 
if it were a quantity. If we are going further, and are to sub- 
ject it to transcendental operations, we must carefully define 
each function that we are going to use, and establish the rules 
which the function must obey. 

The principal transcendental functions are e*, log, and sina, 
and we wish to define and study these when x is replaced by an 
imaginary variable z. 

As our conception and treatment of imaginaries have been 
entirely algebraic, we naturally wish to define our transcendental 


18 INTEGRAL CALCULUS. (ART. 30. 


functions by the aid of algebraic functions; and since we know 
that the transcendental functions of a real variable can be ex- 
pressed in terms of algebraic functions only by the aid of infinite 
series, we are led to use such series in defining transcendental 
functions of an imaginary variable; but we must first establish 
a proposition concerning the convergency of a series containing 
imaginary terms. 


30. If the moduli of the terms of a series containing imaginary 
terms form a convergent series, the given series is convergent. 

Let wp + uy + Ue be Hy eeveee be a series containing imagi- 
nary terms. 

Let 
Uy = Ry(cos®y+ Bee Ue sin®)), uy = 2, (cos ®y+ “| ae sin ®,), &e., 


and suppose that the series 2) + Rk, + R,+-- + By fees is 
convergent; then will the series w+ w+ Ug + be convergent. 

The series ft) + 2, +--+ is a convergent series composed of 
positive terms ; if then we break up the series into parts in any 
way, each part will have a definite sum or will approach a defi- 
nite limit as the number of terms considered is increased in- 
definitely. 

The series wv + wy + Uy +e Uy foveee can be broken up into 
the two series 


Ry cos®, + Ry cos®, + Ry cos ®, + +--+ -- BR, cos ®, + ++ (1) 


and 


V—1(Rsin® + Rsin®, + R, sin, + -....+R,sin®,+----). (2) 


(1) can be separated into two parts, the first made up only 
of positive terms, the second only of negative terms, and can 
therefore be regarded as the difference between two series, each 
consisting of positive terms. Each term in either series will be 
a term of the modulus series Rp + R, + R,+- multiplied by 
a quantity less than one, and the sum of n terms of each series 
will therefore approach a definite limit, as increases indefi- 
nitely. The series (1), then, which is the abscissa of the point 
representing the given imaginary series, has a finite sum. 


Cuap. II. ] IMAGINARIES. 19 


In the same way it may be shown that the coefficient of Nieag 
in (2) has a finite sum, and this is the ordinate of the point 
representing the given series. The sum of n terms of the given 
series, then, approaches a definite limit as m is increased indefi- 
nitely, and the series is convergent. 


31. We have seen (I. Art. 133 [2]) that 
2 3 4 

2 ee BB echo EY cole pa SF a 1 

eh ets is Fa 


when 2 is real, and that this series is convergent for all values of x. 
Let us define e7, where z= «+ yV—1, by the series 





3 a4 
Bee et ah ne arth carte aaererian ina [2] 


This series is convergent, for if z= r(cos ¢ + \/—1. sin ¢) the: 
series 


made up of the moduli of the terms of [2] is convergent by 
I. Art. 133, and therefore the value we have chosen for é? is a 
determinate finite one. 

Write «+ ¥V—1 for z, and we get 


erty /—1 — tee eRe eee yy GE es ry [3] 


The terms of this series can be expanded by the Binomial 
Theorem. Consider all the resulting terms containing any given 
power of x, say #; we have 


Berle (WV 12 N= 1)? yN—1)" 
eet (VT GN 1) CUN rod ivy oe 
ae 2! 3! da eld )s 
or, separating the real terms and the imaginary terms, 

a 2 ia Wee 

a ( oi + loiter = ) 


20 INTEGRAL CALCULUS. fART. 32. 


uP pee ea | 
or a (cosy + V—1.siny), by I. Art. 134. 
Pp! 
Giving p all values from 1 to « we get 
zty/—1 Le as 1 ee aan ay" x MS 
; Gees ane) Taos a ae 
= e (cosy + V—1.siny), [ 4 | 


which, by the way, is in one of our typical imaginary forms. 
digs Olin i, 
we get e’Y-! = cosy + V—1.siny, 


which suggests a new way of writing our typical imaginary ; 
namely, er 4: 
r(cos¢ + V—1. sing) = reev=1, 


32. We have seen that 


+y/-1 __ WSJ=1. 
Cries = ee ; 


let us see if all imaginary powers of e obey the law of indices; 
that is, if the equation 
ere eerae [1] 
is universally true. 


Let U=y+Y, V—1 and v=a%+ Yo </lit 
then e™= em +%~=1 = e% (cosy + V—1.siny), 
e” = e% + Ye Y¥-1 = e%2(COS Yo+ /—1.sin Yo) s 
eve” = et e% [cos (yy + yo) + V—1.sin(y + y) | 
= e%+% [eos(y, + yo) + V—1. sin (gi + ye) | 
— ott, + (y+ Y2)a/-1 
= ett? 


and the fundamental property of exponential functions holds for 
imaginaries as well as for reals. 


EXAMPLE. 


Prove that a"a” = a"t” when w and v are imaginary. 


Cuap. II.] IMAGINARIES. 21 


Logarithmic Functions. 


38. As a logarithm is the inverse of an exponential, we ought 
to be able to obtain the logarithm of an imaginary from the 
formula for e?+’¥-1. We see readily that 


z= 17 (cosd+V—1. sing) = eet oY 77, 
whence logz =logr+¢V—1; 
or, more strictly, since 
Z2=Pr[cos (dy) +2n7) + V/-—1.sin(d)+ 2nz7)], 
logz = logr + (dy + 2n7) V—1 [1] 
where 7 is any integer. 


Ifz=2-+-yV—l,r= s/o? 4+- 4, and d= tan! ; 
a 


whence logz = flog (a? + y?) + V—1. tan“, [2] 


Each of the expressions for log z is indeterminate, and repre- 
sents an infinite number of values, differing by multiples of 
2rV—1. 

This indeterminateness in the logarithm might have been ex- 
pected a priori, for 


etl — cos2a4+V—1.sin2r7=1, by Art. 31. 


Hence, adding 27 ~V/—1 to the logarithm of any quantity will 
have the effect of multiplying the quantity by 1, and therefore 
will not change its value. 


EXAMPLE. 


Show that if an expression is imaginary, all its logarithms are 
imaginary ; if it is real and positive, one logarithm is real and 
the rest imaginary ; if it is real and negative, all are imaginary. 


22 INTEGRAL CALCULUS. [ART. 34. 


Trigonometric Functions. 
34. If 2 is real, 


3 5 














; z Z z ! 
ane seo teeee [1] 
ea ae i: [2] 
cosz=1 — aA eS Gl Nise ia 
by I. Art. 134. ine 
If z=r(cos?+ V—1.sing), 
8 
the series of the moduli, 
phat rth pee ly a 
31751 TI Acre “ 
9 g4 6 
na 2 | ar 41 ss 6! ar is ie: ? 


are easily seen to be convergent; therefore if z is imaginary, the 
series [1] and [2] are convergent. We shall take them as defi- 
nitions of the sine and cosine of an imaginary. 


EXAMPLE. 


From the formulas of Art. 31, and from Art. 34 [1] and [2], 


show that 


zV/—l 


e = cosz + V—1.sinz, 


= Gey ° 
and e-*Y-1 — cosz — V—1.sinz, for all values of z. 


35. From the relations 


e~-l— cosz + V—1.sinz, 


e-*“-1 — cosz — V—1.sinz, 
LY | =Zz4/=1 
GF e 
we get COS 2 = —— : [1] 
i er 1 grav al 
sin 2 = ———____—_ 


Pe ae 7] 


for all values of 2. 


Cuap. II.] IMAGINARIES. 23 


Let a UY acs 
cos(wtyV —1)= aa 
by Art. 34, Ex., 


ey Pe ira 


ae 


a 





ey ey earned , 
= COosx Sa — /—1.sinz 


In the same way it may be shown that 





__ (cosa V—1. sinx)eY— (cosa — V —1.sina)e 
2V—1 


y —¥ = Pee gee 
ol a me 4+ /—1. cosa” LD [4] 


. . 
_— 


sin(a+y V—1) 








If z is real in [1] and [2], we have 


ja /—1 Sant aL 
: ah Ot eum 20N 
COB fe es 
ar/—1 Sow ol aba +! 
‘ ; — € 
Bae ee aT 


If z=yV—1, and is a pure imaginary, 








eee y ae 
cosy V—1 = * = ; [5] 
A a alee Sah RI 
sinyV—1l= —V—1; [6] 


2 


whence we see that the cosine of a pure imaginary is real, while 
its sine is imaginary. 
By the aid of [5] and [6], [3] and [4] can be written: 


cos (+ yV—1) = cosxcosyV—1—sinasinyV—1, [7] 


sin (x + y V—1) =sinwcosy V—1+ cosxsinyV—1. [8] 


24 INTEGRAL CALCULUS. [ART. 36. 


EXAMPLES. 
(1) From [1] and [2] show that sin?z+ cos?z = 1. 


(2) Prove that 
cos (u+ v) = cosucos” — sinwsin », 


sin (w+ v) = sinucosv + cosusiny, . 


where wu and v are imaginary. 


The relations to be proved in examples (1) and (2) are the 
fundamental formulas of Trigonometry, and they enable us to 
use trigonometric functions of imaginaries precisely as we use 
trigonometric functions of reals. 


Differentiation of Functions of Imaginaries. 


36. A function of an imaginary variable, 
2=a“2+y ees. 


is, strictly speaking, a function of two independent variables, 
x and y; for we can change z by changing either x or y, or both 
«andy. Its differential will usually contain da and dy, and not 
necessarily dz; and if we divide its differential by dz to get its 


derivative with respect to z, the result will generally contain ao 


which will be wholly indeterminate, since « and y are entirely 
independent in the expression «+yV—1. It may happen, 
however, in the case of some simple functions, that dz will appear 
as a factor in the differential of the function, which in that case 
will have a single derivative. 


37. In differentiating, the V—1 may be treated like a con- 
stant; for the operation of finding the differential of a function 
is an algebraic operation, and in all algebraic operations “/ pout 
obeys the same laws as any constant. 


Cuap. II.] IMAGINARIES. 25 


EXAMPLE. 
Prove that d(a?/ —1)=2aV—1. da; 
and that dV—1. sina = V—1. cosa.da... 
We have, by the aid of this principle, 
if g=a+tyvV—1, 
| dz= da +~V—1. dy; [1] 


if z2=r(cos¢+V—l1.sind), 
dz = dr(cos¢ + V —1. sing) + rdb(— sing +~/—1.cos¢) 
= (dr +rV—1.d¢) (cos$ + V—1. sing). [2] 


38. Let us now consider the differentiation of 2”, e7, logz, 
sinz, and cosz. 


Let z=r(cos¢+ V—l1.sin¢d), 
then Rie 
2” = 7"(cosmd + V—1.sinmd¢), by Art. 24 [1]; 


dz” = mr" dr(cosmd + V—1.sinmd) + mr"d¢ (— sinm¢ 
+ /—1.cos mo), 

dz” = mr”! [cos (m—1) 6 + V—1.sin(m—1) 4] (cosd 
+ V—1.sindg)dr 

+ mr” [cos (m—1) 6+ V—1.sin(m—1)¢4] (cosd 

+V—1.sin¢) V—1.d¢, 

dz" = mr"-! [cos (m—1) 6+ V—1. sin (m—1) | (dr 
+ rV—1.dd) (cosd + V—1. sin ¢), 


Caen") dz, [1] by Art. 37 [2], 
dz™ a 

— = me" 2 

dz [2] 


and a power of an imaginary variable has a single derivative. 


26 INTEGRAL CALCULUS. (ART. 39. 
39. If z=a+yv—1, 
e* = e* (cosy +V—1.siny), by Art. 31 [1]. 


de* = e*da(cosy +V —1. siny) + &(—siny 
+~V—1.cosy) dy, 


Ff Lei rmenil ou (cosy +vV—1. siny) (de +V—1, el 


de* = e*dz, [1] 
dle” | 
— = €*. 2 
= [2] 
EXAMPLE. 
Show that da? = a? log a.dz. 


40; If z= r(cosd+/—1. sind), 
logz = logr + dV—1, by Art. $8, 
dr+7 a ean te dd 


" 


Moge = +V—1.d¢= 


(dr+r -/—1.d¢) (cos d+ Ae, sie 





dlogz= 
r(cos + V—1. sind) 
dlogz = ©, [1] 
z 
dlogz. 1 
2 = -, 2 
dz Z ba 


ev-1 ene 1 


41, sinz= ———____, by Art. 35 [2], 
ae ee 2] 
z/—l pgin/al + eee 
A eit oes ne neo Oa ce 
9/—1 
P/N ay ga | 
= — de, by Art. 35 [1], 


dsinz = cosz.dz. [1] 


Cuap. II.] IMAGINARIES. 27 


ev -l +- gv al 


cos 2 = ——————____—_- 
2 9 
za/=1 __ Saale eA ns 2/71 __ pT ee | 
ee id dz, 
2 2V—1 
dcosz = — sinz.dz. [2] 


42. We see, then, that we get the same formulas for the dif- 
ferentiation of simple functions of imaginaries as for the dif- 
ferentiation of the corresponding functions of reals. It follows 
that our formulas for direct integration (I. Art. 74) hold when x 
is imaginary. 


Hyperbolic Functions. 
43. We have (Art. 35 [5] and [6]) 


eis oy ee - om 


and ty Cee aa v=, 


e” te Ca 





is called the hyperbolic cosine of 2, 


xe bier: 


where & is real. 


e 
and is written cosha; and 9 is called the hyperbolic sine 





of x, and is written sinha; 





sinha = 22 = — /—1.sinaV—1, [1] 
2 

cosha = © +2" = cosa V—I. [2] 
2 


The hyperbolic tangent is defined as the ratio of sinh to cosh ; 
and the hyperbolic cotangent, secant, and cosecant are the re- 
ciprocals of the tanh, cosh, and sinh respectively. 

These functions, which are real when 2 is real, resemble in 
their properties the ordinary trigonometric functions. 


28 INTEGRAL CALCULUS. [Arr. 44. 


44, For example, 


cosh? « — sinh?a = 1; [1] 

2 x -2Q¢“« 

for cosh? t= Gite SNe 
4 

PH dh aN —22% 

and sinh? = Mian ni 
4 
EXAMPLES. 


(1) Prove that 1 — tanh? = sech’a. 

(2) Prove that 1 — ctnh’x« = — esch’a. 

(3) Prove that  sinh(#+y)= sinhxcoshy-+ coshesinhy. 

(4) Prove that cosh(x+y)=coshacoshy + sinhasinhy. 
een ow iM Pada owe 


45. dsinhs = d—__—_ = Xv, 
2 2 


dsinhaw = cosha.da. 


EXAMPLES. 


(1) Prove dcosha = sinhw.dz. 


dtanha = sech?2.da. 


dctnh a = — esch’a.da. 
dsecha = — sechatanhw.da. 
desche = — eschactnha.da. 


46. We can deal with anti-hyperbolic functions just as with 
anti-trigonometric functions. 

To find dsinh7'a. 

Let u = sinh 12, 
then “x= sinhu, 


dx = coshu.du, 


CuaP. Ii.] IMAGINARIES. 29 


dx 
du = - 
cosh u 





coshu = V1 + sinh? wu, by Art. 44 [1], 


) coshu — V1 + 2, 











: v 
dsinh-!a = _ de aed 
V1 +2 
EXAMPLES. 
Prove the formulas 
dcosh"lw= SSA, 
Var? —1 
dtanh a = Ma daa 
1— 2 
dsech-!z = — seas 
aVv1 — a 
‘ eye Meee rae | 
a Var +1 


47. The anti-hyperbolic functions are easily expressed as 





logarithms. 
Let u = sinha, 
, Pest Car 
then ie ese se SUNY 6 == ; 
1 
24=e"——, 
eu 


ey eee 

e" — 2xe"= 1, 

ee" —2ve"+r=1+4 2", 
e —e=tvV1i+e’, 


e=xtvV14+2'; 


30 INTEGRAL CALCULUS. [Arr. 48. 


as e” is necessarily positive, we may reject the negative value in 
the second member as impossible, and we have 


e= e+ V1+ 2, 
u=log(# + V1 + 2), 
or sinha = log(a+ V1 +a”). [1] 
EXAMPLES. 
Prove the formulas 


cosh—!a = log(# + Va? —1). 


tanh“'x = $log 


sech~'x = log é +f 3 -- ) 
BTcNSE 


‘ 
“y* 


Ge 

pean er ages, 
esch ‘a = log(/-+,/,—+1). 

a Na 


48. The principal advantage arising from the use of hyper- 
bolic functions is that they bring to light some curious analogies 
between the integrals of certain irrational functions. 

From I. Art. 71 we obtain the formulas for direct integration. 




















dx pea: 
=e 81h ae [1 
Vi — 
dx 
= tan— x. 2] 
pas an—' w [2] 
dx my 
—__——— = sec “a. 3 
J t/q? — J a 
From Art. 46 we obtain the allied formulas : 
ee = sinha = log(#+ V1+4 2’). [4] 
V1 + a 
{Fes = cosh to = loge + VFI), [5] 
Vo? — 1 | 


Cuap. IT.] IMAGINARIES. 31 














dar , 1+" 
= tanh 42= 41o . ( 
4 irae ty Mel a 
— (B= seoh te = 109 (4 +45 -1) [7] 
avi —2 a al 


; ~ da a 1 1 \ 
4) ech te=log(e ++) ° 8 
eaetmay Ve wv ) LI 





bo 


32 INTEGRAL CALCULUS. (Arr. 49. 


CHAP TER iT, 
GENERAL METHODS OF INTEGRATING. 


49. We have defined the integral of any function of a single 
variable as the function which has the given function for 7s 
derivative (I. Art. 53); we have defined a definite integral as 
the limit of the sum of a set of differentials; and we have shown 
that a definite integral is the difference between two values of an 
ordinary integral (1. Art. 183). 

Now that we have adopted the differential notation in place of 
the derivative notation, it is better to regard an integral as the 
inverse of a differential instead of as the inverse of a derivative. 
Hence the integral of fv.dx will be the function whose differ- 


ential is fv.dx; and we shall indicate it by dp fe.dx. In our old 
notation we should have indicated precisely the same function by 
f fx; for if the derivative of a function is fx we know that its 


differential is fx.da. 


50. If fx is any function whatever of x, fx.dx has an integral. 
For if we construct the curve whose equation is y = fv, we know 
that the area included by the curve, the axis of X, any fixed 
ordinate, and the ordinate corresponding to the variable x, has 
for its differential ydx, or, in other words, fv.dx (I. Art. 51). 
Such an area always exists, and it is a determinate function of a, 
except that, as the position of the initial ordinate is wholly arbi- 
trary, the expression for the area will contain an arbitrary con- 
stant. Thus, if 7x is the area in question for some one position 
of the initial ordinate, we shall have 


fede = Fx +0, 


where C is an arbitrary constant. 


Cnap. III.] GENERAL METHODS OF INTEGRATING. 33 


Moreover, Fx -+C is a complete expression for | fx.dx; for if 


two functions of x have the same differential, they have the same 
derivative with respect to #, and therefore they change at the 
same rate when x changes (I. Art. 38) ; they can differ, then, 
at any instant only by the difference between their initial values, 
which is some constant. 

Hence we see that every expression of the form fx.dx has an 
integral, and, except for the presence of an arbitrary constant, 
but one integral. 


51. We have shown in I. Art. 183 that a definite integral 
is the difference between two values of an ordinary integral, and 
therefore contains no constant. Thus, if Fv+C is the integral 
of fx.dz, 


b 
ap ‘fe.du = Fb — Fa. 
In the same way we shall have 
b 
f fede = Fb— Fa; 


and we see that a definite integral is a function of the values 
between which the sum is taken and not of the variable with 
respect to which we integrate. 


Since f, feeder = Fa— Fb, 
b 
fe b 
[fede = — { fede. 
b a 
EXAMPLE. 


Show that (g : fu.da + f fede = J “fecd. 


02. In what we have said concerning definite integrals we 
have tacitly assumed that the integral is a continuous function 
between the values between which the sum in question is taken. 
If it is not, we cannot regard the whole increment of Fx as equal 


34 INTEGRAL CALCULUS. [ArT. 53. 


to the limit of the sum of the partial infinitesimal increments, 
and the reasoning of I. Art. 183 ceases to be valid. 


1 
Take, for example, f ed 
a 


-—l 
fea fe ated er —%, by I. Art. 55 (7) ; 


and apparently 


f Se we EN aN ee 
-1 a U) p=1 uv fn 


1 
But f ne ought to be the area between the curve y = 5 the 
-1 x 


axis of x, and the ordinates corresponding to 7=1 and a= —1, 
which evidently is not —2; and we 





Y 
see that the function 5 is discon- 
tinuous between the values x= —1 
and a=1. 


aS hala The area in question which the 
definite integral should represent is 
easily seen to be infinite, for 


[Batt a fk 
1 @ € ie ‘ 


and each of these expressions increases without limit as € ap- 
proaches zero. 


53. Since a definite integral is the difference between two 
values of an indefinite integral, what we have to find first in any 
problem is the indefinite integral. This may be found by in- 
spection if the function to be integrated comes under any of the 
forms we have already obtained by differentiation, and we are 
then said to integrate directly. Direct integration has been illus- 
trated, and the most important of the forms which can be in- 
tegrated directly have been given in I. Chapter V. For the sake 
of convenience we rewrite these forms, using the differential 
notation, and adding one or two new forms from our sections on 
hyperbolic functions. 


CuaP. III.] GENERAL METHODS OF INTEGRATING. 30 








fsine.de = —cos2. 
feos v.dx = sinw. 
fianz.de= — log cosa. 


_fetne.dz = log sine. 


f da . poe | 
— SI W@W. 
V1 — x 




















at | sinh ae = log (a4 +-V1 + a"). 

V1 + 2? 

= eogh-'2 = log (a +Va?—1). 

We? —1 
if; oe == tan! a. 

1! + x 

dx i 1 

7 2 = tanh ‘w= ylog- ai 
faesre* 

ea aa 


dx Laat Cy AAC, u te 
Sayre tan sech-*a = log(L +45 1), 


dz 1 recs 
ee ewe t= — log PAE ee 1). 
Snes “\e - 


= = vers! a, 
v2 xe — x? 


36 INTEGRAL CALCULUS. [ART. 54. 


54. We took up in I. Chap. V. the principal devices used in 
preparing a function for integration when it cannot be integrated 
directly. 

The first of these methods, that of integration by substitution, 
is simplified by the use of the differential notation, because the 
formula for change of variable (I. Art. 75 [1]), 

f “u= if uD,x becoming | udx= uw ay, 
Pune / dy 
reduces to an identity and is no longer needed, and all that is 
required is a simple substitution. 


(a) For example, let us find { ae V1+ loge. 
dx 
Let 1+log¢x=2; then care dz, 
and { GviF boge= 2dz=22!=2(1 + loga)?. 
When, as in this example, a factor of the quantity to be 
integrated is equal or proportional to the differential of some 


function occurring in the expression, the substitution of a new 
variable for the function in question will aor a the 














problem. “ 
b) R qd . 
(6) Require a — 
LOhGac ee then e’dx=dy, 
de _ ede __ dy ’ 
eet ea eee 4 
and. f Os | ds a a tan-!y = tan—e, 
e+e” 1+/ 


(c) Required | secx.da. 





Cuap. III.] GENERAL METHODS OF INTEGRATING. 37 


Let z=sinz; then dz=cos2.da, 


cos’4 = 1 — 2’, 








cos «.dx f dz is Veen al 
—_— = J ——_.=4log—** by Art. 53 
if cos?a ene oo , 
fseoa.dx = plog E522 _ log tan(=+ =). 
sin x re 
EXAMPLES. 
Prove that (1) | cscav.dz = flog pic: COS log tan <- 
1+ cos% 2 
4 /1 — 2 
(2) oak i ee 
V1i— ¢ 2 


Suggestion: Let «= cosz. 


0d. The formula for integration by parts (1. Art. 79 [1]) 
becomes 


{ude = uv — | vdu, [1] 
when we use the differential notation. It is used as in I. Chap. V. 


(a) For example, let us find | x" loga.dza. 











Let u=loga, and dv=x"dz; 
then dit= so 
) x 
n-+1 
and v=— ) 
n+1 
+1 n n+1 
fe log x.dxz = - loga — dx = — log @ ee. 
. nN n+1 n+1 n+1 


(0) Required | x sin-*2.dz. 


Let u=sin™g2, and dv=xdx; 
dla 


then du = ———., 
Vi-we 





— 


38 INTEGRAL CALCULUS. 


and Vv 





2 2 
fasine.de = 7 sina — +f. < 
2 Ne 


[ART. 55, 


fasinte.dx = 9 sine +4(cos“!a + av 1 — 2). 











(c) Required “hese aia 5° 
Let wa Ee": 
and Ailes seers 
(1 +2)" 
then du = (xe* + e*) dx = e*(1 + x) da, 
and v= — 2 3 
1+2 
xe* dx re z 
Pde cca edzy= — 2 = ee 
(1+)? Site er prac 1+ a 
EXAMPLES. 
Q) f= = sin S22, 
Vi— 8% iat V138 
(2) {| xtan-'a.da= ; += tan-'x — 4a. 
dx 1 1 
3 {= >- Side Oe i ha aa 
() (1 — 2)? PLAID TOES 5 
dx x 
(4) | =- — V2ae — 2? + aversé' 
V2 axe — 2 
(5) {via 2. dx = ——% e seen + < sin 2=9 an 
a 


Suggestion: Throw 2ax — - into the form a? — (a — a)?. 


(6) 1+ cosx 


Br dx = log(# + sina). 


Cuap. III.] GENERAL METHODS OF INTEGRATING. 39 


(7) e+ SMe 9, ¢tan ne 
1+ cos2 2 
Suggestion : Introduce ; in place of x. 


1 
eas (nm — 1) (log art 


(9) (eee dx = loga [log (logx) —1]. 


(8) 


a) (a= = > ztanz + logcosz, wherez = sin!z. 


pee | =). 
hd ie tone 3 75 estan (5+ 5) 


uh (12) sinvdax _ log(a+bcosx) , 
a+b cosx b 


«a3 {o-“>- “1 (@ 4.2), 
(13) edn 48 tan~' (x + 2) 
(14) oe = Tog 1+ 

1—2 

3 da 1 et — 3 

\ ae Oe Lom h 
(15) {5 BG a0 Cz ) 


dx r b 
i a tT tan }- 
Ge d?cos?a+ b?sin*a ab ( ) 








40 INTEGRAL CALCULUS. [ART. 56. 


GHA PDE Resive 
RATIONAL FRACTIONS. 


56. We shall now attempt to consider systematically the 
methods of integrating various functions; and to this end we 
shall begin with rational algebraic expressions. Any rational 
algebraic polynomial can be integrated immediately by the aid of 


the formula 
an +1 
fe de = : 
n+l 


Take next a rational fraction, that is, a fraction whose nu- 
merator and denominator are rational algebraic polynomials. 
A rational fraction is proper if its numerator is of lower degree 
than its denominator ; improper if the degree of the numerator 
is equal to or greater than the degree of the denominator. Since 
an improper fraction can always be reduced to a polynomial 
plus a proper fraction, by actually dividing the numerator by the 
denominator, we need only consider the treatment of proper 
fractions. 





57. Every proper rational fraction can be reduced to the sum 
of a set of simpler fractions each of which has a constant for \a 
numerator and some power of a binomial for its denominator ; 


that is, a set of fractions any one of which is of the form 


fac 


Let our given fraction be . 
x 





(e—a)™ 


If a, 6, c, &c., are the roots of the equation, 
FL), (1) 
we have, from the Theory of Equations, 
Fa = A(x — a) («#—b)(%—c) «+ (2) 


Cuap. IV.] > RATIONAL FRACTIONS. 41 


The equation (1) may have some equal roots, and then some of 
the factors in (2) will be repeated. Suppose a occurs p times 
as a root of (1), 0 occurs g times, ¢ occurs 7 times, &ce., 


then Fa = A(x — a)? (x — b)*(a@— cc)" + (3) 
Call A(a—bd)% (a — Cc)" = ha; 
then Fa= (%— a)? dx, , 

Fa Sa 


ft fe tA ENE Riga HEI ne 


d eee ee ee 
- Fe (@—ayoe @—ay ga (@—a) $a 
J Sa 
di alla al 
he = a)? 7 (2 — a)? bau 
Sa 
gf he oa pu 
eee ge is a new proper fraction, but it can be reduced 


to a simpler form by dividing numerator and denominator by 
* — «a, which is an exact divisor of the numerator because a is a 
root of the equation 


pa Sat tga 
fe ae oe = 0. 


If we represent by fx the quotient arising from the division 


of fe — ‘ px by «—a, we shall have 
a 








Ja 
fo ga fie 
Fu («—a)?’ (x—a)? dx’ 
where aoa is a proper fraction, and may be treated 
— (#—a)? de 
precisely as we have treated the original fraction. 
ha 
H SE ac I 
CT) ar CEE) ea CET ar 
By continuing this process we shall get 
oe 2 Ae Sea Jp-14 
fx pa pa pa a Pe pa 4 Sp& 


Fu (@—ay G@—aPa! @—a) Re a 


42 INTEGRAL CALCULUS. (ART. 58. 


In the same way si can be broken up into a set of fractions 


having («7 —b)?!, (e—b)*", &c., for denominators, plus a frac- 
tion which can be broken up into fractions having (#—c)’, 
(a—c)’-)...-, &c., for denominators; and we shall have, in 


the end, 





fe A, Ay ioe A B, 
Fx (ea)? @—ayat ery (x — 6)! 
By B 
ae Gob) obo esaes + ae: mee + Kk, [1] 


where is the quotient obtained when we divide out the last 
factor of the denominator, and is consequently a constant. More 
than this, A must be zero, for as (1) is identically true, it must 


0 
be true when «=; but when x= ©, i becomes zero, be- 
x 


cause its denominator is of higher degree than its numerator, 
and each of the fractions in the second member also becomes 
zero; whence AK = 0. 


58. Since we now know the form into which any given rationat 
fraction can be thrown, we can determine the numerators by the 


aid of known properties of an identical equation. 
3x2—1 


(« — 1)’ (@+1) 


Let it be required to break up into simpler 
fractions. 

By Art. 57, 
(abo s 
(e—1)?(@+1) (w—1)? w@—-1 9 w+ 

and we wish to determine A, B, and C. Clearing of fractions, 
we have 
8e—1= A(#+1)4+ B(w—1)(#+1)4+C(#—-1)’. (1) 
As this equation is identically true, the coefficients of like 
powers of x in the two members must be equal; and we have 
B+C=0, 
A—2C0=3, 
A—B+C=-1; 


Cuap. IV. ] RATIONAL FRACTIONS. 43 


whence we find A= 1; 
aes hs 
C= —1; 
ne 82 —1 Teta “4 @) 





(x —1)?(@+1) = (z—1)? me neta 


The labor of determining the required constants can often be 
lessened by simple algebraic devices. 

For example; since the identical equation we start with is 
true for all values of x, we have aright to substitute for x values 
that will make terms of the equation disappear. Take equa- 
tion [1]: | 


3%—1=A(e#+1)+ B(x+1) feety Ce OGEie Pay 


bet eo 1, 2-2 A, 
eo N, 
then 2xe—2= B(x+1)(#—1)4+C(a—-1)’; 
divide by x —1, 2= B(w#+1)+C (a—-1). 
here =, Pex 2B: 
enh. 
then —x+1=C(x#-1), 
C= —1. 
EXAMPLES. 


(1) Show that when we equate the coefficients of the same 
powers of a on the two sides of our identical equation, we shall 
always have equations enough to determine all our required 
numerators. 
9a? + 9x — 128 


(2) Break up eB GEET) 


into simpler fractions. 


59. The partial fractions corresponding to any given factor 
of the denominator can be determined directly. 


44 INTEGRAL CALCULUS. (ArT. 59. 


Let us suppose that the factor in question is of the first degree 
and occurs but once; represent it by «—a. 








ae eke fix 
Fause <a wae (1) 
by Art. 57, where | 
ee Fx 
ie eon 
so that Fa = (a — a) $a. 


Clear (1) of fractions. 
fe= Apu+(a—a) fix. (2) 


As (1) is an identical equation, (2) will be true for any value 
OL %:.\ Let a=, 


Ja = Ada, 
yee a (3) 


a result agreeing with Art. 57. 

Hence, to jind the numerator of the fraction corresponding to 
a factor (x —a) of the first degree, we have merely to strike out 
Srom the denominator of our original fraction the factor in ques- 
tion, and then substitute a for x in the result. 

If the factor of the denominator is of the nth degree, there are 
n partial fractions corresponding to it. Let («—a)” be the 
factor in question. 





FE Ay A, ide Ya 
Fs (@—ayt @—ay* @—ay't "tena 
where Fa = (% —a)” pa. 


Multiply (4) by («— a)”, and represent (a — a)” ‘ by ®a. 
x 


bx = A, + A,(w — a) +A; (x — a)? + «+» + A, (ae — a)" 


WX, a\n 
ATEN a)”. 


Cap. IV.] RATIONAL FRACTIONS. 45 


Differentiate successively both members of this identity, and put 
«=a after differentiation, and we get 


A, — da, 
Aa ©'a, 
1 
A, = a1 Oa, 
A, == S Oa, 
eocce 5 
ee aM? (n-1) gy 
" (n—1)! 


Although these results form a complete solution of the prob- 
lem, and one exceedingly neat in theory, the labor of getting 
the successive derivatives of ®x is so great that it is usually 
easier in practice to use the methods of Art. 58 when we have to 
deal with factors of higher degree than the first. So far as the 
fractions corresponding to factors of the first degree and to the 
highest powers of factors not of the first degree are concerned, 
the method of this article can be profitably combined with that 
of Art. 58. 


60. As an example where the method of the last article 
applies well, consider 


30—1 cae! B» C 
e(e —2)(¢ +1) - x hee orl 
3x2—1 1 
A= — =-, 
(cca vay iar 2 


32—1 o 
B=| ———— =-, 
eal: 6 i 


8a —1 bee te 
“tess le 8” 


32—1 rd 
x(x —2)(%+1) 





Tat 
eM uar x—2 32+1 


46 INTEGRAL CALCULUS. (Arr. 61. 


Again, consider 
Dee af) ea) ae ei oe nea 


Ee care 2S ene wat 
— 1 fp=-V=1 


B=| 1 | OC 0 OR ais nee 
ae.) bile st aay are eS eres pemnenrs PS ana =!) 
gulA/ET eva Oa eed 2 


1 AP ELT 1 A est 1 2] 


sem a 








61. Let us now consider a more difficult example, where it is 
worth while to combine our methods. 
bts sacha gh 
(@—1)*(@ +1) 

w+1l=(*#+1)(2?—x#+1) 


= (“#+1)(e«—4—4V—3)(@—44+4-V-3), 


To break up 








PR ASL) 2st ee Se 
(©—1)*(@?+1) (a#—1)*(@+1)(@’?—aw+1) (#—1)? 
A, A, A, B C 
ah Gpeliaye (@—1)? ml 241 gages 
BD 
ie Ee 4 1 
a—t+3$V—3 2 


a facia area | m1 
* (x 


ae eae | ] as 
(pH) (ee ae ‘ 


sulcpences ss « 
dalemnmernyearere 
ae bessrereraty —1)*(@+ Nearer ye ri 


Cuap. IV.] RATIONAL FRACTIONS. 4T 


“ia si 
3 "i 3 _ 1, Q#-1) 
Pee Nee fe PEN 68) ee ed 


Substitute these values and clear (1) of fractions. 
24 (a?+1)=24(a+1) (2’—a+1) +24 A,(a—1)(e+4+1) (2’—a2+1) 
+ 24.A, (a —1)?(@ +1) (a —x% +1) + 24.A, (w —1)? (@ +1) 
(a’—a+1)+ (a—1)*(a@’—x+1) —8(2e—1) (x—1)*(@ +1) ; 


152°—51 a +452'+ 62? —512°+45¢ —9 = 24A,(@—1) (a +1) 
(a? — +1) + 24.A, (2 —1)?(@ +1) (2? —2 +1) +244, 
(@—1)? (+1) (@+2—-1),. 
The second member of this equation is divisible by 
(@—1)(*+1)(a’?—x2+1), 
therefore the first member must be divisible by the same quantity. 
Dividing, we have 
15a? — 36%+ 9= 24 A, + 24 A;(~@ —1) + 24A,(@ —1)?. 


Let ¢=1, —12=24A,, 
1 
A; = x 9” 
and we vet 


15a? — 862-421 = 24.A,(# —1) + 24.A,(a—1)2. 


Divide by «—1; 
15a —21 = 24A,+ 24A,(x—1). 


Let «= Ls, —6= 24 A,, 
1 
A,= aha 
15% —15 = 24.A,(a—1). 
Divide by x—1; 15 = 24 A,, 
yy gga 


* 


48 INTEGRAL CALCULUS. _ (ART. 62. 


Hence 
wd ee 
(a —1)* (2 +1) (e@—1)*) 2: (e@—1)! | 4 (2 Se 
Lien. 1 1 1 
ae Se ee (2) 
e+1 8 «—4¢-—4V—-3 8 2—-444V—8 


62. Having shown that any rational fraction can be reduced 
to a sum of fractions which always come under one of the two 


forms 





=u and nei it remains to show that these forms 
—a)” x—a 


can be integrated. 
To find = is Adx 


a a)” 
let Z2=%t—a, 
then Wa == (a, 
(@—a)”™ —a)” 2" (n—1) gn (n—1) (a—a)" 
To Ls 
x— a 
let Z2=2—A, 
then d= 00, 
and ao = Af 2= Alogz = Alog(#— a). [2] 
r— z 


Turning back to Art. 58 (2), we find 


_ (8a—1) da _ ={=., a ae du |. (édaaeenan as 
(x —1)?(a#+1) (%—~ t—1 J oti eg 
x—1 


addy 








+ log(« —1)— log (w# +1) = ———_ ce ona es 
x — 


Turning to Art. 60 (1), we have 


Mrcmsycenwiu a) 


= tloga + $log(x — 2) — $log(a +1). 








Cuap. IV.] RATIONAL FRACTIONS. 49 


63. If imaginary values come in when we break up our given 
fraction, they will disappear if we combine our results properly 
after integrating. 

We know (Art. 28, Ex. 2) that if the denominator of our 
given fraction contains an imaginary factor, («—a—bV—1] 1p ie 
it will also contain the conjugate of that factor, namely, 
(w—a+bdV—1 1)". Moreover, since by Art. 59 the numerator 
of the partial fraction corresponding to (7 —a—bV —1)" will be 
the same rational algebraic function of a+b —1 that the nu- 
merator of the Le fraction corresponding to (x —a+bV—1) 








is of a — bV—1, these two numerators must be conjugate imagi- 
naries by Art. Oy Ex. 3. Hence, for every fraction of the 
A+Bv—1 
form ———__———____ we shall have a second of the form 
G@—a—bV—1)" 
A-—BV=-1 
(eon 4-6 —1)" 
oD Se (Ase Ba) pak 
(eo —b.\/—1)" (n—1) (e—a—bV—1)" 
by Art. 62 [1]. 
es eh (A BN 1)" 
(2—a-+b~V—1)* Cone Lye wpb — 1) 
beth) (2a oV—1) = xX+YV—1, 
X and Y being real functions of x; 
then (ex—-a—bvV—1)"'=X-—YvV— 
Beer BV ( A BNA! gn 
(w—a—bV—1)" (2 —a+bV—1)" 


1 ay a Barr 1 (Ao 1) 
Pee VY ti (nl). XPV 
1 _@AX—2BY) 1] 


ee arenes |g 


a result which is free from imaginaries. 


~ 


50 INTEGRAL CALCULUS. (ART. 63. 


If ik y 
we have the pair of fractions, AEB RES : “ae A> Byes : 
cg | pg aul 2—atovV—1 


aver, ec. 
x—-a—bv—l 
by Art. 62 [2], 


AA BNL n= (A— BV—1) log(e—a--bV—=1); 
Se Dead Pedal 

















log (a — a— b V—1)= 4log [(@ — a)’ + 6] — V—1. tan” — e a 
log(#—a+bV—1)=4log[(w@—a)? +0] + V—1 stant 2 
Hence Wee ee da + f A = BN eee a dx 
e—a—bV¥—1 Re A Sa | 
= Alog[(#—a)?+0?]4+2 Btan™ e , [2] 
—a 
which is real. 
The form of [2] can be modified by adding a constant. 
= + tan“ > = 5 tctn 10 0 etn) 2 Se 
2 x — A 
Hence Alog[(«—a)?+ 67] +2 Btan ora [3] 
differs from [2] by the constant Bz, and therefore ‘is a true 
value of A+B Neate ae Au RATS 4 — BNon dae 
LA aeany eae Yn aN “a 


Turning back to Art. 61 (2) we find 


eater o= {ae al Bes Fela 


lx da dx dx 
+8f( 4% yf —— “2 ___ - {——"= 
“J a—1 x1 ®, ae . e—$4+hV—3 © 


Olja oie on 1 
ES) 1)8 ud Geet Fer pis 1) 


+ 54 log (aw +1) — flog (&’ —x%+1). 


Cuar. IV.] RATIONAL FRACTIONS. 51 





























EXAMPLES. 
ON ae Bes de = 2 + log®=?. 
> @) Jaga et toes 
ee Co} es ss tans 2. 
Oss ee co oe 
(6) aie ae 
(7) aE = Hog +E tant 
8) fe = Hoe eh 
O Sep ery™ Te te 
+ dtan'e — ay + Hog(e? +1). 


(10) ee aie e—xV2+1 
+1 aa etaV2+1 


WF —— [tan (a V2 +1) + tan-! (aV2—1)]. 








Saar ee OV es 1 24/2 


(11) ae e+ av2+i 1 tan (2X2) 


o2 INTEGRAL CALCULUS. [ArT. 64. 


CHARTER: 
REDUCTION FORMULAS. 


64. The method given in the last chapter for the integration 
of rational fractions is open to the practical objection that it is 
often exceedingly laborious. In many cases much of the labor 
can be saved by making the required integration depend upon 
the integration of a simpler form. ‘This is usually done by the 
aid of what is called a reduction formula. 

Let the function to be integrated be of the form w”-!(a+bz"), 
where m, 2, and p may be positive or negative. If they are in- 
tegers, the function in question is either an algebraic polynomial 
or a rational fraction; if they are fractions, the expression is 
irrational. ‘The formulas we shall obtain will apply to either 
case. 


Denote a+ ba" by z; then we want § w"-!z?dza. 


Let P=U 
and ode = dv, and integrate by parts. 
du = pe dz = bnpx"-! 2-1 da, 
or aa 
m 
fam teds a TE OND (ym tn gel cl, [1] 
m m 


This formula makes our integral depend upon the integral of 
an expression like the given one, except that the exponent of x 
has been increased while that of z has been decreased. 

We get from [1], by transposition, 





a 2, m 
forte tde = Fo a ale dx. 
np np 


Cuap. V.] REDUCTION FORMULAS. 58 


Change m-+ 7 into m and p—1 into p, whence m is changed 
into m —n and p into p+1, and we get 


m=—n »p +1 
alee dag ae (ym n-1 3p $1 
J bn(p+1) bn(p+1) i fate te 


- a formula that lowers the exponent of x while it raises that of z. 
Since 2 bar, 


Zz? = 2-1(a + ba"), 
hence 


(meg dx = (eter + ba") dx = a fant dx 


+ b feria dx ; 
therefore, by [1], 


xz? bn i: 
— TEP | atl eplda = a | a2 da + b fantntr-'de, 
m 


Mm 
fone Id¢z= ene On ENP) f‘gntartget dz. 
am am 





Change p into p+1. 


ifr dx = am apt = Bem Fp EO) (amen-terde, [3] 
aM am 
i 


Change m into m— n, and transpose. 


f gm —ler dx = ARS _ a(m—n) glo dae. 4) 
b(m+np) b(m+np) 


We have seen that 


fore de x af amar td +. b (ante ter tda, 


and, from [1], 





am eP om 
b fi am te-lep-l de = —— par eda; 
np mp 


54 INTEGRAL CALCULUS. [ART. 65. 


hence 
a ae m 
forede = af amar tde + — — 2" ae, 
np np 


foe de a am 2 ga LAD fate lde. [5] 


m+np m+np 





Change p into p+ 1, and transpose. 


fam tede= se uaa aha au lassi itl oo halts arcane 
an(p+1) an(p+1) 








Formula [3] enables us to raise, and formula [4] to lower, the 
exponent of w by n without affecting the exponent of 2; while 
formula [5] enables us to lower, and formula [6] to raise, the 
exponent of z by unity without affecting the exponent of a. 


Formulas [1] and [3] cannot be used when m = 0; 
formulas [2] and [6] cannot be used when p= —1; 
formulas [4] and [5] cannot be used when m =—np; 


for in all these cases infinite values will be brought into the sec- 
ond member of the formula. 


Bali eke z=a+ ba, 


and our last four reduction formulas become 


mw~p+l " 
fen Pda —_ Ned = b(m+p+1) am Pda. [3] 


am an 


am—tept+) ~a(m— 1) 
m—1 .p Er ng te EN Bs Rieu ES Fgid m—2 ~p - 
fe Su b(m+p) b(m+>p) eae he 











= x” 2? a x 
fi SA fh Ee ek on” 5) a Es [5] 
m+p m+p.e 


feed sail naar at. MEPHL ( pm-1y9+1 Gy, [6] 
a(p+il) a(p+l1) 


If m and p are integers, and m >0 and p>0O, a repeated use 
of [5] will reduce p to zero, and we shall have to find merely 


the if el dz. 


Cnap. V.] REDUCTION FORMULAS. ay) 


If m<0 and p>0, [3] will enable us to raise m to 0, and 
then [5] will enable us to lower p to 0, and we shall need 


se dae 
only 
x 
If m>0 and p<0, [6] will raise p to —1, and [4] will then 
lower m to 1, and we shall need f aa 
Zz 


If m<0 and p<0, [6] will raise p to —1, and [3] will raise 


m to 0, and we shall need 


am 
fen Om —, 
m 


dx 


—=log2, 

x 

da dx 1 

—= | —— =- log(a+ dz), 

Z is Help" h Bre) 

dx olf dx Lr a+ ba 
a= og aE ES aS 

XZ x(a + ba) a 2G 


Hence, when n =1, and m and p are integers, our reduction for- 
mulas always lead to the desired result. 


EXAMPLES. 
+ba , BF b? b 1 
1) (= — Slog" ai eee 
O) o(a+ba) Ge ax Qe Zea das 
(2) Consider the case where n= 2, rewriting the reduction 


formulas to suit the case, and giving an exhaustive investi. 
gation. 





lah mh x 
(3 ae kp eeraayan Sablac be) 


1 b 
Pio Bade. sf | Aap. 
 s(ab) om a 


be. 4 
¥ 
ff 


56 INTEGRAL CALCULUS. [ART. 66. 


CHAPTER VI. 
IRRATIONAL FORMS. 


66. We have seen that algebraic polynomials and rational 
fractions can always be integrated. When we come to irrational 
expressions, however, very few forms are integrable, and most 
of these have to be rationalized by ingenious substitutions. 

If an algebraic function is irrational because of the presence 
of an expression of the first degree under the radical sign, it can 
be easily made rational. 

Let f(a, Va + ba) be the function in question. 








Let z=Vat be; 
then 2" = at be. 
nz" dz = bdx, 
nz”) dz 
da = ——— ; 
b 
—a 
= e 
b 





Hence fie. Va + bx) da = AC - ae ae de 
which is rational and can be treated by the methods of Chapter IV. 


EXAMPLES. 


(1) Mee dx = +4 ./a+4log(/e—1). 


Z a hte NCO Oa 
eta di aN 
(3) f[ev@ + a)++/(@ +a) |da 


_ny/(@+a)th navy (%+ayr* _ 
2n+1 n+1 bi Oa 


Cuap. VI.] IRRATIONAL FORMS. 5T 


67. A case not unlike the last is f. fla, See bas. 


Let z2=Vet+Vatba; 
2=c+ Va-t ba, 
(2"—c)"=a-+ bz, 


(2"—c)™"—a 
x = ie eee ae oe ee wer he 


b 
dx — Mn (z"— c)m gn} dz 
b 
Hence Ah me 
fre. Ve+ Vat ba) dx 
_ mn i Sa TF ie E =H, 2 |e Gy Rig da: 
EXAMPLES. 
(1) Find {—-— ae 
Ve+ Vat ba a 


+ (2) Fina f_—__@ __. 
View 1s 


68. If the expression under the radical is of a higher degree 
than the first the function cannot in general be rationalized. 
The only important exceptional case is where the function to be 
integrated is irrational by reason of containing the square root 
of a quantity of the second degree. 


Required if. S(a, Va'+ bx + ca?) da. 


First Method. Let ¢ be positive; take out Vc as a factor, and 
the radical may be written VA + Ba + 2’. 


Let VA+ Be + e?=autz, 
A+ Be t+e=27’ + 2a2+ 2’, 
Pe: | 


Cis 





B—22 


58 INTEGRAL CALCULUS. [ART. 68. 


2(2° — Bz + A)dz 
(B—2z)?° 
ee eae ape 
VA + Bu ied Ny Ne ee cree 
B—2z 
and the substitution of these values will render the given func- 
tion rational. 


dz = — 


b) 


Second Method. Let c be positive; take out Ve as a factor, 
and, as before, the radical may be written VA + Ba + 2°. 


Let VA+Ba+e=./A+ 22; 
A+Bu+er=A+2/A.cz+ 02 


_2/A.2—-B 
t= 
oe 
a alti MP 
Pe 


and the substitution of these values will render the given func- 
tion rational. 

If c is negative the radical can be reduced to the form 
VA + Bu — 2, and the method just given will present no 
difficulty. 


Third Method. Let c be positive; the radical will reduce to 
VA + Bu+2?. Resolve the quantity under the radical into the 
product of two binomial factors (a —a)(#—f), a and f being 
the roots of the equation A + Ba +27=0 





Let V(x —a)(%@ —B) =(e@—a4)z; 
(%—a) (% —B) = (w—a)?*2*, 
ye de 
1-—? 
ee 22(B— a)dz 
a—#) 


V@= 4) @—B) = (@—a)2= Ba 


9 
pied 


Cuap. VI.] IRRATIONAL FORMS. 59 


and the substitution of these values will make the given function 
rational. 

If c is negative the radical will reduce to VA + Bx — a’, and 
may be written V(a— 2) («—) where a and B are the roots 
of a — Bu — A=0, and the method just explained will apply. 

In general, that one of the three methods is preferable which 
will avoid introducing imaginary constants; the first, if c > 0; 


a 
the second, if <0 and ——>0; the third, if e<0 and— <0. 


a 
If the roots a and £ are imaginary, and 4 = ig is negative, it 


will be impossible to avoid imaginaries, for in that case 
A+ Be —~2x’ will be negative for all real values of 2. 


69. Let us compare the working of the three methods just 


da 
iven by applying them in turn to the example { ————————_. 
g y applying p Migr wy ¢ 
lst. Let V2+3a+e?=x+2; 
(—_— = (282. | Oye eee 2 dz 
ee (3 —2z)? 2—32+2 3—22 
= — log(3 — 22), 
SS = B+ 2 22 FB TH) 
V2 4+-3a +2 
Beppe OAl2 + Bae 
loo — BAe t2V2 +3042" 
oa 12a +428 — lle —4e 
= log [3 +2a+2V24 3x +427]. (1) 


2d. Let V2 8 ear = 4/2 + az ; 
{—. — beri 2 —32-+/2)dz 1—2 
= a — 2)? /2.2—32z+/2 


=2 (4, a og (Art. 53) 





\ 


60 INTEGRAL CALCULUS. [ART. 69. 
(- dx ux Neches Ahad tel UP cate bina 
JV243a4+@ — ge a, pian amar, 


oo te +2avV2 +30 +0°+2432+27—2 
E. t+ 2./2.¢0¢+2—-2—8¢—a 


pe ate V24+38a+2 
ip 2 L/S B IE Meninie 


= log(3 + 247+2V2+ 3e+27) — log(2,/2—3), 
or, dropping the constant log(2./2 — 3), 





f = oe 8 +20 +2 V2 + 8a FH). (2) 
V24+3a+2 


3d. Let eae * Gai) a 


























2f —2zdz 1-—2z 1+2 
NEES SER NOLS ? ; = log r} 
[ees = (1 — ee 
2 
1 ae 
f dar mag aS Nero log V@+1+ Va+? 
V2+3a+2 i A Ve+1—V2+2 
«+l 
og ttl +2VI4F3eF P+ x42 
a+l—a-—2 


= log (8 +2a+2V2+432+4 a?) +log(—1), 
or, dropping the imaginary constant log (— 1), 


S re tet VERE, (3) 
EXAMPLES. 

(1) f= US a ES V4+2e—V2—a@ 

(2+32)V4—2% 4-2 oi Loe Ve 


dx 





== log (4-+%+Vat-+a). 





®) (Tz 


iad ahs Tego a+ ba + ca? 
(3) eet s/c is om hae Vag ia ) 


Cuap. VI.] IRRATIONAL FORMS. 61 


70. If the function is irrational through the presence, under 
the radical sign, of a fraction whose numerator and denominator 
are of the first degree, it can always be rationalized. 


Required f. of (« ‘ aa da 











la +m 
Let 4= Qin b 
le + m’ 
gn +b 
le + m’ 
b — mz” 
i Wen gp tee Uo 
lz” —a 
dy — Ram — bd)z" ‘dz 
aa: 
and the substitution of these values will make the given function 
rational. 
EXAMPLE. 





Ge Nie a ee 


71. Ifthe function to be integrated is of the forma”-!(a+ba")?, 
m, n, and p being any numbers positive or negative, and one at 
least of them being fractional, the reduction formulas of Art. 64 
will often lead to the desired integral. 


EXAMPLES. 
1) [EB = tein te 2 RE (3 +208), 
dx gag tlie eae V1 — 2? 
ve /1 — a ut 2 a" 
f x dx ADR a d x 
(2 — a7)2( = + — 30° Ne 
x i Semmes & ee ) ey ys: ass 2a 


(=, x da af) _ 20? +32") | 
(a? + x)3 3(a> + x*)2 


\ Fi 
X& 














62 ' INTEGRAL CALCULUS. [ART. 72. 


72. We have said that when an irrational function contains a 
quantity of a higher degree than the second, under the radical 
sign, it cannot ordinarily be integrated. It would be more cor- 
rect to say that its integral cannot ordinarily be finitely expressed 
in terms of the functions with which we are familiar. 

The integrals of a large class of such irrational expressions 
have been specially studied under the name of Elliptic Integrals. 
They have peculiar properties, and can be expressed in terms of 
ordinary functions only by the aid of infinite series. 


Cuap. VII.] TRANSCENDENTAL FUNCTIONS. 63 


CHAPTER VII. 
TRANSCENDENTAL FUNCTIONS. 


73. In dealing with the integration of transcendental functions 
the method of integration by parts is generally the most effective. 


For example. Required | x(logx)’dz. 


Let u = (loga)’, 
dv = 7.da; 
2 log x.dx 
du = , 
x 
2 
v=-—) 


f2(log2)* = ees — rloge.de = < [ (log w)?— loga + 4]. 


Again. Required | e’sinw.dx. 
u = sin, 
dvu=e da; 
du = cos x.dx, 
T=.6" 


fe sin 2.da = e*sin x — | e*cosx.da, 


fe cos x.du = e*cosx+ | e*sin 2.dz; 


e* (sina — cosa) 
cee APR Sasha 
2 


e* (sine + cosx 
and fe0os v.de = aaa 


whence f esin x.dv = 


64 INTEGRAL CALCULUS. (ART. 74. 


EXAMPLES. 
& fod ; 3d(logx)? 
(1) SHG Se Bel oe 1 [Oar 
6 log x 6 
(m + Lye (m +1)3 





(2) ae on log (1 —@). 
(3) | e™/(1—e’*) .da= - [va —e*) 4 since 


74. The method of integration by parts gives us important 
reduction formulas for transcendental functions. Let us con- 


sider f sin” v.da. 
Miss SINT. 
dv = sina.dx ; 
du = (n—1)sin"-*x cos x.da, 
vV=—cOsxv; 


fs" a.da = — sin""'2 cosa -+(n—1) | sin*-?a cos’?a.da 
= —sin"“xcosx+(n =1) { (sint*e — sin” x) dx ; 
som 1 . n—l n—1 . n—2 
sin” v.dxz = — = sin""!x cosx + —— | sin"—?a.da. [1] 
n n 
Transposing, and changing n into n + 2, we get 


fi sin” v7.dx = J 
« n+1 


In like manner we get 








sin"*+!a cosa +” ae sin" ¢.da.) [2] 


La n—1 
f costa.de = -sin 2 cos*“1 4 4+ ——— 
n 


f cos" x.dx2 = — 1 
n 


If n is a positive integer, formulas [1] and [3] will enable us 
to reduce the exponent of the sine or cosine to one or to zero, 


cos"? x.dx, [3] 





;sinwoos' 4 42 cos" *?a.dx. [4] 


Cuap. VII.] TRANSCENDENTAL FUNCTIONS, 65 


and then we can integrate by inspection. If » is a negative 
integer, formulas [2] and [4] will enable us to raise the ex- 
ponent to zero or to minus one. In the latter case we shall need 


f liad , or if pte which have been found in Art. 54 (c). 
COS & sin x 





EXAMPLES. 


3 a 
sin? a + 4 +4. 


4 Sa sinz cosa 
(1) § sinte.de = — ———— ( 2 


- . 8 
sin x cos®x 
(2) | cos°a.da = ————— (costa + 


ee 
— (sina cos2+ 2). 
6 \+H¢ i®) 





(3) Hey bo COS & 


+ log tan~. 
sina 2sin?a 7. 


(4) Obtain the formulas 





P Tees n—1 j 
f sink x.dx= — sinh"-!x%coshxz — sinh”~?a.da. 
nv if) ory 


fsinb'ede= J aihive esate oe { sinh’*?2.do. 
n+1 n+l. 

















1 . VL tis iad 1 9 
fcoshte.de= — sinh x cosh”"!a-+- —— | cosh”-*2.dz. 
n n 
7 Nae ntl, ~ e+2 eal 
cosh" x.de= — sinh x cosh"t"a-+—— 4 cosh"**2.da. 
' n+1 n+1 
Keabit) 
(5) Ee pee 4] Beene Ly 
oe sinh? x cosha + | 


75. The (sin-'x)"dx can be integrated by the aid of a reduc- 
tion formula. 
Let : = sing; 


then v= sing, 


dx = cos2z.dz, 


and f(sinrayrae= {2 cosz.de. 


66 INTEGRAL CALCULUS. [ART. 76. 


Let Wa, 
dv = cos2z.dz ; 
dus na dz, 


v= sinz; 


fe cos z.dz = z"sinz — n fzr-sin 2.0z. 
« 


Ne ‘sinz.dz can be reduced in the same way, and is equal 
to —2z""*cosz +(n—1) f? 2-2 cosz.dZ : 


hence 


fe cosz.dz = z"sinz + nz""!cosz — n(n —1) fz cosz.dz, [1] 
or f(sint2)"de = a(sin“!2)"+ nV1 — ¢(sin7}a”)"" 


—n(n— 1) f (sinz)"*ae. [2]. 


If n is a positive integer, this will enable us to make our re- 
quired integral depend upon | dz or } sin-'w.da, the latter of 
which forms has been found in (I. Art. 81). 


EXAMPLES. 


1) Obtain a formula for | (vers7!x)” da. 
(1) 


(2) f (sin? 2)tde = of (sin a)\—4.8.(sin2)?-4+4 3.2.1] 
+41 —o?sin-'«[ (sin 2)?— 3.2]. 


76. Integration by substitution is sometimes a valuable method 
in dealing with transcendental forms, and in the case of the trigo- 
nometric functions often enables us to reduce the given form to 


an algebraic one. Let it be required to find if (fsin2) cos xv.da. 


. 


Let Z= sing, 


dz = cosx.dx ; 


f (fain x) cos x.dx = {fede 


Cuar. VII.] TRANSCENDENTAL FUNCTIONS. 67 


In the same way we see that 


{ (feos) sinx.dx = — f fede if z=cosa”, 


and 


f[i(eine, cosa) ] cosa.da = {Ure Vi—2*)|dz if z=sing, 
JL feose, sin 2) | sinzdx=— f [7 V1—z*) | dz if z=cos2, 


or, more generally, 


f f(sine, cosx) dx = {se Meters if z=sing, 


1-~7" 


1; S(cosx, sina) da =—{ F( SDM pak if z=cos”. 
V1—2? 


Since any trigonometric function of « may be expressed in 
terms of sinw and cosa, the formulas just given enable us to 
make the integration of any trigonometric function depend on 


the integration of an algebraic function, which, however, is 
frequently complicated by the presence of the radical V1—2’. 


77. A better substitution than that of the last article, when 
the form to be treated does not contain sing or cos# as a factor, 


; % 
1s 2= tan-- 
| 


BE 2 dz ° 
This gives us dx = ——.,, 
1+2z 
22 
sinx = ———,, 
1+2 
4 om 
cos @ = 53 
z 








LPG? 
whence facing, cos x) dx = 2ft(» eae bee [1] 


As an example, let us find if: Rg PEt 
a+bcosx 


68 INTEGRAL CALCULUS. [ART. 78. 


Here we have 


f dar =2 {— dz =2f | dz 
a+b cosa ial a+tb+(a—b)# 


1+? b 
( +2) a+b, ae 


2 dz 2 me ) 
ae ——__—__— = —__—__- tan +2 
a—b sak 2 Ve—Pe a+b 

a—b by I. Art. 77, Hixaae 


dx 2 a—6 Nee 
Hence (= 1 3(/ -tan= |, if b. 
Ja+tbcosx /g2— 3? ar a+b a5) ach" 


78: afi sin”x cos" v.dx can be readily found by the method of 














Art. 76 if m and n are positive integers, and if either of them 
is odd. Let be odd, then 


cos" =cos""xcos a = (1 —sin 2a) E “cosa, 
if sin” xcos"v.da = | sin" x(1 — sin?) 7 cos x.da. 
Let Z== Sins, 
dz = cos 2.da, 
ie 
fosinne cos" v.da = | 2"(1 — 2") 2 dz, 


which can be expanded into an algebraic polynomial and ‘inte- 
grated directly. 
If m and » are positive integers, and are both even, 


n 
f sin"& cos" @.dx = f sin” a” (1 — sin?@) 2da. 
n 
sin” (1 — sin?x)2 can be expanded and thus integrated by 
Art. 74 [1]. 
If m or n is Racy: and odd, we can write 
cos"” = cos""*acosx, or sin™x=sin™ 2 sing, 


and reduce the function to be integrated to a rational fraction 
by the substitution of 
= COSY, Or 2% =—sin7w. 


f sin” x cos"x.dx can also be treated by the aid of reduction 


formulas easily obtained. 


Cuap. VII. ] TRANSCENDENTAL FUNCTIONS. 


ia f tan"xdx and (“= oe 





69 


can be handled by the methods 


of Art. 78, but My can tie simplified greatly by a reduction 


formula. 
We have 


fran 2.de = {tan tan’x.dx = {tan ?e (sec? — 1) dx 


See — f tan"-*2.d2, 



































n”- ly 
whence fitan 2.de = —— = — f tan” xd 5 [1] 
a f da _ (ten _ ons ={- (tana) _ da 
tan” xv tan” x ; tan" & tan”—2a° 
whence ft cs i tee —f. sot ‘ [2] 
tan” (n— 1) tan”"*e tan” 7a 
EXAMPLES. 
é 10 8 
if Pires’ a: cos’ z.d¢ = Aes DNs FA enka 
10 8 
(2) | cos’a Vsina.da = cece Sa 
x (3) (Sas x.da nN! A eae Ne ey 
V Cos x v 
; , a Sane 
eee toes sin'c.dq = ——— | 
2 3 12 16 
~ 5) eee = seca + log tan =. 
sin #@ cos’ a 2 
da. COs a o x 
6 ee = secx — —log tan-- 
(6) sin’ x cos’ x 2 sin? en Ps 8 2 
1 
7 =— log sina. 
x 7) ae Were eye aioe 


T0 INTEGRAL CALCULUS. [ART. 79. 


1 Vb +a+ vba. tan 5 
AS) ore == —____ lor —__ ae 
eee J? — oF Vb-fa — Vb=a. tan® 
Soe 9 4+5 tans 
GVoni eh ek ara — “tan-} : 
( ig baa ee 3 a 3 
Pal ain ae oe ~ + s1op(se-e 
(0) Soreee rage 5 0s sing Ob COR arts og (3 +2 cosa) 





(1 1) fee con ade _5 sine — sytem" (5 tan} 
(5 +4 cosa)? 95+4cosx 27 3 2 


| (a—c) tan 5 +b 
ae : tan! 


12) SS oe 
oe a+b sina-+ecosa Ve—eP—e L Vae—b—e 


Cuap. VIII. ] DEFINITE INTEGRALS. fie: 


CA Acre tak VL ET. 
DEFINITE INTEGRALS. 


80. In I. Art. 183, a definite integral has been defined as the 
limit of a sum of infinitesimal terms, and has been proved equal 
to the difference between two values of an ordinary integral. 

We are now ready to put our definition into more precise, 
and at the same time more general, form. 

If fx is finite, continuous, and single-valued between the 
values x= a and x= J, and we form the sum 


(a, pad a) fa =f (Xp — 2%) fx +(%— _) fy + coe (Yp1— Xn—2) Sen—2 
me (b By Gu1) fln=15 


where 2, %, V-+++X,_; are n—1 successive values of a lying 
between a and J, the limit approached by this sum as n is in- 
definitely increased, while at the same time each of the increments 
(x, — a), (@ — 2), etc., is made to approach zero, is the definite 


b 
integral of fx from a to b, and will be denoted by f Su.da. 


If we construct the curve y=/x in rectangular co-ordinates, 
this definition clearly requires us to break up the projection on 
the axis Of X of the portion of 
the curve between the points A 
and B into n intervals, to multi- 
ply each interval by the ordinate 
at its beginning, and to take the 
limit of the sum of these products 
as each interval is indefinitely decreased; that is, the limit of 
the sum of the small rectangles in the figure, and this is easily 
proved to be the area ABA,B,. 

Now the area 4BA,B,, found by the method of I. Chap. V., 


is | freee | =| fede] 





12 INTEGRAL CALCULUS. (Arr. 81. 


»b 
Therefore f fe. dei j Fd | — : fee ‘ [1] 
a iO 2=a 


eh ~ 
Shat 1s, if fe.dx is the increment produced in J Ju.dx by 


changing x from a to 0. 

It is to be noted that the successive increments (#,— 4a), 
(x, — 2%), (#;— %), etc., that is, the successive values of da, 
are not necessarily equal; and also, that if we multiply each 
interval, not by the ordinate at its beginning, but by an ordinate 
erected at any point of its length, the limit of our sum will be 
unaltered. (v. I. Arts. 161, 149.) 


81. It is instructive to find a few definite integrals by actu- 
ally performing the summation suggested in the poate 
(Art. 80), and then finding the limit of the sum. 


db 
(a) f v.adx. 


Let us divide the interval from a to } into n equal parts, and 
call each of them dw. 


Then ndz=b—a. 
Our sum is 
S=ada+(a+dax) dx+(a+ 2dx)du+---+(a+(n—1) da) dx 
= nade +(1+2+3+---+(n—1)) dx? 
=a(b—a) shana seed, dx", 


since ndx = 6—a, and the sum of the arithmetical progression 
n(n—1). 
2 


etl ARE we 1) dx? = 4 (n*dx* — ndx*) = (6—a)’ = a)* _ (6—a)da = dex 


1+2+4+3+4-.--4(n—1)= 


Hence 


Cuap. VIII.] DEFINITE INTEGRALS. 73 


As we increase 7 indefinitely, dx approaches zero, and 


b Shas ats a ri 2 2 
Sem aes ol a (8 paces 


Se Re 2 ise 
b 
(0) t eda. 
Let PO dana 





n 
S = e*dax + ett da + ett? da +... - ettin-Dad dy 
ae e* dz [1 + eX + e2ax af esda af. ete -- OO AEE | : 
but 1+e"+e"4+...+e"-)* is a geometrical progression, 
and its sum is 
enix ey | Gs eo-4 28 1 
et Ler 1 et ee, 1 


eee 
eee et «OM ctf Or aE 
Soraen ( ) 





dx 


9 
e* — ] 


Hence S= 





ce limit [ da 
x ats b a . 
and fe dx = (e iy) y ol g i) 


approaches the indeterminate 








but as dx approaches zero, — 
e~ — | 


form rt but since the true value of 


a 1 Ja=0 e* |a—0 
b 
f e* dz = e’ — e*. 
| a 





(c) {) coste.de, 
0 
Let dx =”, and let n be an odd number. 
n 
Then 


S = da + cos*du-da + cos’ 2dx-dx + --- + cos’ (n — 2) dx- du 
+ cos’ (n — 1) da-da 
= dz + cos*dx- dx + cos? 2dxu-da + --- + cos’ (4# — 2dx) - da 
+ cos (x — dx) -du 
= da +.cos*dx- da + cos®2dx-da-+ --- — cos®2du- dz 
— cos’ da da, 


T4 INTEGRAL CALCULUS. (Arr. 81. 


since cos (7 — ¢) =— cos@. 


Hence the terms cancel in pairs, and we have left 


S = dx 
and ff costed = Bee [ae = 0. 
: 
(d) (] sin?a.da. 
0 


Let da = a and let n be an odd number. 
n 


S = sin?0-daw+sin’dx-da+sin?2 da. dx + ---+sin?(n—2)da- dx 
+sin? (n — 1) dx-dx 


=sin’da-da+sin? 2da-da+---+ sin'(5 — 2do)do-+sin'(S —_ dada 


=sin’da-de+sin? 2da-da+---+cos?2dx-dx+cos*dx-da, 





since sin G — $) = cos®¢. 
Then Sade + det dx ="="da, 
since sin? d + cos*¢?= 1. 
Therefore S=*- a 
dene 
2 ss 
and Hi sin’ v.dx2 = —- 
0 4 


(e) { & 


Here it is best to divide the interval between a and 3d into 
unequal parts. 

Let the values 2), 2, %,+++ @,_, be such as to form with a and 
b a geometrical progression. 


"lb 
For this purpose take q = Ne so that ag” = 0. 


Cuap. VIII.] DEFINITE INTEGRALS. T5 


Then the values in question are aq, aq’, ag’-:- ag", and the 
intervals area(q—1), ag(q—1), ag’?(q—1)--- aq” (q—1), 
and the sum 


il ai (qin n-l(g — 
ee) 22(g—1) | ag Ws a ar (q- 1) 
a aq ag ag 


=n(q—1). 

To prove our division legitimate we have only to show that 
each of our intervals, a(qg—1), ag(q—1) --- ag™*(q—1), 
approaches the limit zero as n increases indefinitely. Since 


b 


ime 


the limiting value of q as n increases must be 1, as otherwise 


iy g” would not be finite. 
limit x ae 
Therefore , _.. [ag*(q—1)] = 1 Lag" (¢g—1)]=0. 


We have then 


ies... [S] = mit re 1)] = mT ney 








1 b 
__ limit °F a a 
—1l 
~ g@=1} logg eee a ))> 
b 
since nlogg = log= we 
limit b st 4 —1 b 
But log — = log-—. 
i q=1 ee A os ap Pil sea | °8 G 
For 3 [f = | Sess hea Te 
| Tes | 1 
FY _jq=t 


Therefore if ¥ = logb — loga. 


76 INTEGRAL CALCULUS. [ART. 82. 


EXAMPLES. 


(1) Prove by the methods of this article that 





(2) By the aid of the trigonometric formulas 
cos 6 + cos 26+ cos36+---+cos(n—1)0 
ae 3] sinng etn S —l1— cosnd | 
sing + sin26 +sin36 +---+ sin(rn—1)6 
4 a — cosné) ctn : — sin nd 


I 


b 
prove that f cosxv.da = sinb —sina, 


a 


and Ah sina.dx = cosa — cosb. 
Qr 
(3) Show that ue sin’x.da = 0, 
0 


ud T 
and that if cos? a.da = —- 
0 P 

oetl_ qmti 


, using the method of 
m+1 


b 
(4) Show that { a Toe 
Art. 81 (e). 


82. When the indefinite integral can be found, the definite 
b 
integral | fw.dx can usually be most easily obtained by em- 


ploying the formula [1] Art. 80, and this can always be done 
with safety when /# is finite, continuous, and single-valued 
between «=a and «=D. 

Of course, if the indefinite integral is a multiple-valued func- 
tion, we must choose the values of the indefinite integral cor- 
responding to x =a and «= 6), so that they may be ordinates 


of the same branch of the curve y = f. Su.dx. 


4 


Cuap. VIII.] DEFINITE INTEGRALS. T7 





rel] A 

Consider, for example, J ; ia -- The indefinite integral 
e/-1 Abe 

= ean and tan-tx is a multiple-vaiued function. 


Indeed, y=tan~'wz is a curve consisting of an infinite number 





of separate branches so related that ordinates corresponding to 
the same value of x differ by multiples of +. On the branch 
which passes through the origin, when «= —1, y=tan!a= inh ; 
on the same branch, when x=1, y=tan-!'2=*- On the next 
branch above, when «=—1, y= tan +z = ; and when x=1, 
U fee “2, On any branch, when «= — 1, y= tan ‘a= — ‘i + 7 ; 


and on the same branch, when e=1, y= : + nz. 











H {<< >= tan*(1 — tan-!(—1) = 242=7:; 
ence ae (1) ( ) f ae ; 
or 1 Cae SN See TEM ae 
STR SCORING? | Gay Sammy 
dz T T T 
=-~-+nr—|{ —-— aaien 
[ere roma ( ite) 2 


a b 
By {_ fx.dx we mean the limit approached by of fv.dx as b 
is indefinitely increased. 
EXAMPLES. 


(1) Work the examples of Art. 81 by the method of Art. 82. 


(2 (ne dx OY Cae 


cos?a 


a d 4 ae nal 
ee (V2 — 1). 
ont of aps 


1 eae 
AG) 0 aie hy 20 


78 INTEGRAL CALCULUS. [ART. 83. 





(5) ( © —7 ifa>0, and —7 if a<0, and 0 if a=0. 
w, Vi eee ‘ 2 


(6) ede =! ifa>0. 

0 a 

(7) f e-“sinma.dz =—"— if a>0.. 
0 av +m 


(8) e~™ cos mx.dx ara if a>0. 





[ (9) ( __@ ha 
01+2acos¢+2* 2sind 





(10) ( —_-@___= mca 
0 1+2ecos¢+2*" sind . 


83. When fa is finite and single-valued between «=a and 
«= 06, but has a finite discontinuity at some intermediate value 


e=C 
b c b 
A) fede = f fede + f fre. det, 


b 
and therefore fx.dx can be found by 





Art. 82 when the indefinite integral 
of. fxe.dx can be obtained; but when fx becomes infinite for 
x=a, or for «=0, or for some intermediate value x=c, 
special care must be exercised, and some special investigation 
is usually required. 

If fx is infinite when «=a and “fr.dat approaches a finite 
limit as « approaches zero, this limit is what we shall mean by 


b b 
f fu.dzx; if | fx.dx increases indefinitely as e« approaches 


a+e 


b b 
zero, we shall say that +] fu.d« is infinite; and if ( Su.da 


ee’ a+e 


neither approaches a finite limit nor increases indefinitely as ¢ 


Cuar. VIII.] DEFINITE INTEGRALS. 19 


approaches zero, we shall say that fk fe.dz is indeterminate. 
It is in the first case only that le Jfxu.dx can be safely employed 
in mathematical work. 

If fx is infinite when «=b and he. dx approaches a finite 
limit as e approaches zero, that ait is the value of fb Se.dx. 

If fx is ee nite when «=c, and each of the expressions 
of “fre. dlee and fe. dx approaches a finite limit as e approaches 
Bere: the sum of these limits is f fe.dx. Should either or 
both of the expressions, 

fe dx, J fede, 

fail to approach a finite limit as e approaches zero, fu.dx is 
either infinite or indeterminate, and cannot be safely used. 

When the indefinite integral of fxv.dx can be obtained there 
is little difficulty in deciding on the nature of if Jfu.dx in any 
of the cases just considered, or in getting its value when that 
value is finite and determinate. 


For example, 


1 

AL. +s 

(a) ihe — is infinite, since 
0 


= = log x and ( = log (1) — loge = log=, 
€ 


and Pe tcc indefinitely as « approaches zero. 





(b) f i fas 3 is not finite and determinate, for 


dx 1 1+2 
— log 
Jf Te a nes: 


[+ Slog (as 
mie we 5 eu 


and increases indefinitely as « approaches zero. 














80 INTEGRAL CALCULUS. [Art. 84. 














Eats ; ; : 
(c) f —_——— is finite and determinate, for 
EN [see 
dla Aerie 
Sg Bes 
Var — 7 
itis da . iy CE One| . eee 
—___. = sin — sin7*0 = sin-+'——_.,, 
0 ~/q?2 — o a) a 


F Pacis : BAU 7 
and its limiting value as « approaches zero is sin~1(1) or 5 


a aoe 
(d) di aes, is finite and determinate, for 
— 2 


¥ 
ade 5 2 
iiaieacaee SG AY occ Hg pet 
ee 2 ( ) 5 ( )*s 
1~€ oda 5 2 
PII RAED Fs MEET 3 
Veet ame 


and its limiting value as « approaches zero is 3 — 2. 


+€ 


2 
ada 5 2 
i ae eee ae 
7 ? 
ee Dea ae 


and its limiting value as « approaches zero is —?—3, and 
consequently 


84. When, as is sometimes the case, the indefinite integral 
cannot be obtained, and the function to be integrated becomes 
infinite at or between the limits of integration, it is only neces- 
sary to investigate the limiting value of «f(a+e) as e ap- 
proaches zero if fx becomes infinite when «=a; of ef(b—e) 
if fe becomes infinite when x=0; and of both ef(¢—e) and 
ef(c+e) if fx becomes infinite when x=c. If each of the 


6 
values in question has zero for its limit, Af fu.da is finite and 


determinate, otherwise it is infinite or indeterminate. 


Cap. VIII.] DEFINITE INTEGRALS. 81 


For, in order that the area HE'B'B should not increase 
indefinitely as e¢ approaches 
zero, the rectangle AA'E’'E, 
whose area is ef(a@+e), must 
approach zero as its limit; and 
the same reasoning holds good 
for the other cases considered 
above. 

Let us bly this test to the 
examples considered in Art. 83. 


(a) (ft Pa iecuibecauce ahah | =, 
) ea) € 


nan 
dx . hd hd 
if [et a es indeterminate, for 
ne pa 


limit € __ limit € ETO hit es ity 
ee (he)? |) e= 0) 968 |e = 0) 2 a 


and 
limit € ae: 
e=0)1—(14+ 6)” 


(c) f pie ae finite and determinate, for 
nae la? — a 


vara € |= 220 € |= e207 
Peeevae (a) ) SL eae a} 69 


2 “cl . 
(d) f iene is finite and determinate, for 
0 (1 aa x) 3 














bole 
° 














= 


limitf ¢«(1—e) ale LOR 
and 


ell e(1 +e) ]=- 
<=0l a= +9)! 


Le t Ce fei ay 12 0. 


82 INTEGRAL CALCULUS. (ART. 84. 


Even when, as in the examples just given, the indefinite 
integral can be obtained, there is a decided advantage in using 
the very simple method of this article. For if the application 
of the test shows that the definite integral in question is infinite 
or indeterminate, the labor of finding the indefinite integral is 
saved; and if the application of the test proves the definite 
integral finite and determinate, it follows that the indefinite 
integral does not become infinite for the value of wx which 
makes the given function infinite, and consequently when the 
indefinite integral has been obtained, the method of Art. 82 
can be used without hesitation. 

As an example, where the indefinite integral cannot be ob- 
tained, let us consider at some length 


"(tog)" a 
Or "ie 
f (108) v 


; re Pte 2 ; 
If n is positive, (to is continuous and single-valued be- 
; Ly 


tween «=0 and «=1, but becomes infinite when x=0. We 

must then investigate the limiting value of <(Iog") as € ap- 
€ 

proaches zero. 


(loge) is indeterminate when «=0, but its true value is 
€ 


easily found to be zero if n is positive, whether n is whole or 
1 n 
fractional. For positive values of n, f (108) dx is, then, 
0- x 
finite and determinate. 
If n is negative, call rn=—™M. 
1 n 1 
Then ii (1085) dx ={ = GO 
0 x 0 ( =) 
log — 
x 
1 
(108: 
a 
is continuous and single-valued from x«=0 to x=1, but be- 
comes infinite when «= 1. 


Cuap. VIII.] DEFINITE INTEGRALS. 83 


We must, then, find a Pewume ; 
(r=) 
. ove 
limit 1l—e 


e= 0} 1 m—1 ; 
| m (log ) 
Les, 


if m=. 1, this is zero; if m=1, it is 1; and if m>1, it is 
infinite. 





which proves to be 





* ae 
cc) 


is, then, finite and determinate if m <1, but infinite if m=1 
or m>1; and we reach the result that 


1 n 
f, (108 3) dx : 
0 a 


is finite and determinate if » >—1, but infinite if n=—1 or 
n<—l. 


EXAMPLES. 
(1) Prove that 





o1 
; 1 log loge ee ; oh | ish BN ate 
1—2 a0 ' ® 1—2z 
are finite and determinate. 
(2) Prove that 
a 2m 
if aan oa se where m and n are integers, and 

0 1—a* Jo 1i— 





= 1 

a ° : 

if ; . dx, are not finite and determinate. 
g°l—zex 


v1 
(3) Find for what values of n { (loga)"dx is finite and 
0 


determinate. 


84 INTEGRAL CALCULUS. (ART. 85. 
1 n 
(4) Find for what values of m and n if (log dx is 
0 x 


finite and determinate. 


1 
(5) Show that {° "(1 —2)"~dz is finite and determinate 
e/0 


if m and n are positive. 
(6) Prove that Hf ; log sinw.da is finite and determinate. 
0 


(7) Show that the following integrals are finite and deter- 
minate, and obtain their values: 


i AG 2 hier 
0 fq?—a2 2 
ih. dx 
ae 
0 Jan — a 
ifs GEN ie aan 
1 /e2?_4 3 


85. It was stated in Art. 82 that by f fede we mean the 
limit approached by af; fo.de as b is mndannitely increased, and, 
as we have seen, if ine indefinite integral f. Jx.dx can be found, 
there is no difficulty in investigating the nature of fi fe.da and 
in obtaining its value if it is finite and determinate. ‘there are, 
however, many exceedingly important definite integrals of the 
form Hf ‘fo. de whose values are obtained by ingenious devices 
srittiott employing the indefinite integral, and these devices 
are valid only provided that the integral in question is finite and 


determinate, since an infinite value not recognized and treated 


Cuap. VIII.] DEFINITE INTEGRALS. 85 


as such, or a value absolutely indeterminate, renders inconclu- 
Sive any piece of mathematical reasoning into which it enters. 

If we construct the curve 
ae 20, fi fe.dx is the limit- 
ing value approached by the 
area ABB,A,, as OB, is in- 
definitely increased; and in 
order that this area should 
be finite and determinate, it is clearly necessary that the area 
BCC,B, should approach zero as its limit as OB, and OC, are 
indefinitely increased, however great the amount by which OC, 
exceeds OB,. 

That is, limit 


b= @ | if fet 
6 


c= @ 


Y 





must be equal to zero no matter how much more rapidly ¢ in- 
creases than 0. 


86. The investigation of the limiting value of fotke, as b 


b 
and ¢ are indefinitely increased, is usually made with the aid of 
the following important theorem known as the Maximum- 
Minumum Theorem. 


If a given function of x is the product of two functions, one 
of which v does not change its sign between x =a and x=b, 
and if M is algebraically the greatest and N the least value of the 


b 
other factor u between x =a and x=b, fi uv.dx lies between 
b eb a 
—M { v.dx and N{ v.dx. 


To prove this theorem, let us first suppose that v is positive 
between «=a and w2=b. Now, M—vu is positive for the 
values of « considered, (Z— wu)» is positive, and therefore 


b b b 
Jf Or=ujv.de > 0 and Mf v.de> f wo.de. (1) 


86 INTEGRAL CALCULUS. TART. 86. 


u— WN also is positive for all values of « between 2=a and 
x=b, (uw—N)v is positive, and therefore 


b b b 
(u—N)v.de>O0 and uv.dx > Nf v.de. (2) 


b b b 
Hence Jf wv.de lies between Mf v.de and Nf de. 


It is easy to modify this proof to meet the case where v is 
negative. 


(a) As an example of the use of this theorem, we will prove 
ali “eda finite and determinate. 

e-** is single-valued, finite, and continuous for all positive 
values of «; if, then, we can show that uf ew dx has the limit 


zero as b and ¢ are indefinitely increased, c remaining greater 
than 6, our proof is complete. 


, at ol 1 ee 
e-@* can be written 2’e-” . —, and — never changes its sign. 


As a increases xe” eventually decreases, and continues 
to decrease toward zero as x increases indefinitely, as may be 
proved by determining its value for «= o. 


We have aa =| oy | — j =| = 
Pr Beers ye ee 


Hence, eventually the greatest value of x’e-* between = bd 
and a=c is b’e-’, and the least value is c’e—. 
Therefore, by the Maximum-Minimum Theorem, 


a 6492 : 1 Oe a fe 
be uk Wad Pea cau epll Cine ae for 2 = 
Gre b Hb 6. Dae 


se) (roe a 
Ie d be 


As 6 and ¢ are indefinitely increased, the first and third mem- 
bers of our last inequality approach the same limit, zero. Con- 











sequently the limiting value of f e-** da is zero, and 
b 


Crap. VIII.] DEFINITE INTEGRALS. 87 


oO 
(: e-~°dx is finite and determinate. 
2/0 


oO . 
Sina) &>. 
az. 
x 





(6) Let us consider it 
0 


sin aa 
x 
and continuous for all positive values of a. 


is equal to a when «= 0, and is single-valued, finite, 





By integration by parts, 


























sin aa cosaz 1 COs ax 
i AGL: a 
xv aa Ce A 
Therefore 
| cea d See cosac 1 fh cos ae 
2= 
a/b v b a0” 
: : : cos ab cos ac 
As 6 and ¢ are indefinitely increased, and 


approach the limiting value zero. 
1 
— does not change sign between x=b and w=c, and the 
ea 

greatest value of cosaa is 1, and its least value is —1. 


rents (Ss dx Si eee r>—1f"" {a 


x 11, cosas Ren: 
b’-¢ b ye Pras 


COS ax 
and the limiting value of (ee dx as b and ¢ increase is 
°° sinaa 











dx; and therefore 





zero. ‘The same thing is true of i 





sin aa ; : : 
he dz is finite and determinate. 
0 x 


(c) uf; cos(2*)dx is finite and determinate. 
0 


For cos (2) is single-valued, finite, and continuous for all 
positive values of x; and it is not difficult to prove that 


if C08 (2”) dx approaches the limit zero as b and ¢ are indefinitely 
b 


' increased. 


88 INTEGRAL CALCULUS. [ART. 87. 
We have 
c c a na ne ; pe c Ci 
ih cos (@) da = 210 CORI faae +4f sin (2) da, 
b b 


q 2% 2% |, ae 
sin(c’) _ sin(0”) 
2¢ 2b © ata (ie 
Be A NE a sin (") 
creased indefinitely is zero; and the limiting value ot f, da 


oe 


The limiting value of as 6 and ¢ are in- 








can be proved zero by the method followed in (0). 


EXAMPLES. 


sin aw 





(1) Construct the curves y=e™; y= 5. Of Sx. COS (ar ee 


(2) Prove that the following integrals are finite and deter- 
minate : 


foe) .* 9 e fee) . [eo ° 
sin’ x sing e sin mx 
0 00° 0 a/ 0 ny 


wv 


co ice) 2) 2 a2 
=) ey) TS 
Al e-°“ cos ba. da, ay e7a gm | daz, uF e” @. da, 
0 0 


0 
log (5 + } - dx. 


1 
0 e* — 1 








(3) Show that ali xe-*.dx is finite and determinate for all 
0 


values of » greater than —1. 


87. When we have occasion to use a reduction formula in 
finding the value of a definite integral, it is often worth while 
to substitute the limits of integration in the general formula 
before attempting to apply it to the particular problem. 


ag". ae 
For example, let us find f —_———» 
. 0 Va? — a 


We can reduce the exponent of w by [4], Art. 64, 


if} gla? doh Lame geth a (m—2) (mn 
b(m+np) b(m+np) 


Cuap. VIII. ] DEFINITE INTEGRALS. 89 


For our example this becomes 


é m~26 2 p2\h 2 sk 1 
fe(@e—ay de = an (aay? a*'(m—2) a8 (a? — 2”) *da. 


—m+1 —m+l1l 
When «=0, and also when w=a, dBA Nock Ad re 
—m+1 


Hence 


fea — #7) 4da = a'(m — 2) =) fo3(a! — x)? da ; 
0 m—1 Jo : 


fee- a?) —* dar oe . a — a)? dx 
0 0 
Lee : 3 : at {a 2 923 dar 
Ga) ah 0 
is . 3 . 4 . a® r dx 
G42 0 /q?— a 
Therefore sree Oye sd z 3 A ° : ux 
BE Nis rd hg ee ay Ob) a 
EXAMPLES 
fo) {aia =2.2 a, 
0 Va? — a? 3.0 
9 ; 2_y?.dax Piyerar: 
(2) f Va = x oF 
e Ta AE 1 7a' 
a] eva? — 2? - dae =— - ——. 
@) f, # va ne 
re 3 Dabs 
4 ge? (a? — x?)?-da=—-. —.- ra*. 
(4) [ a(@—2) ain 
(5) Jf, sin a.de By devotes Mt 2). when ais even, 
0 BE Ae celt r 
Pre AGT (nie 1) 


yA a hl a) when n is odd. 


90 INTEGRAL CALCULUS. [ARrT. 88. 


iy 


(6) Show tat f? cos” ©.dx ={- sin" xv.da. 


(7) ee = _ 1.3.5...02n—1) 
2.4.6...21 2 


Suggestion: let x= sind. 





9 faite, 2.4.6...20 
ge 93.0.7...(2n+1) 


(9) From Exs. 7 and 8 obtain Wallis’s formula 
mw 2.2.4.4.6. e O.Gyes 


2" BES By 79.0 


1 gn-l dx 1 a2" dae 1 n+l dx 


oVit 0 V1 a eee 


Suggestion : 





b 
88. When in finding {_ fx.dx the method of integration by 


substitution is used, and y =F is introduced in place of x, we 
can regard the new integral as a definite integral, the limits of 
integration being Fa and Fb, and thus avoid the labor of re- 
placing y by its value in terms of « in the result of the indefinite 
integration. 


Let us find ff. eV1 — e*. da. 
Substitute y =e", 
dy = ae“ daz. 


Hence fie Toe. dx= 1 ( VI=#. dy. 
When x=—o, y=0, and when z=0, y=1. 


Therefore [ete wey! on Lf vi=¥ a dy 


There is one rather rare class of cases where special care is 
needed in using the method just described. It is when y has a 
maximum or a minimum value between x=a and x=), say 
for «=c, and & is a multiple-valued function of y. 


Cuap. VIII.] DEFINITE INTEGRALS. \ OL 


For suppose y a maximum when 2 =c, then as oo ‘increases } 


from a to b, y increases to the value Fe, and then decreases to™= 


the value Fd, instead of simply increasing or decreasing from 
Fato Fb. If wis a single-valued function of y, and ¢(y)dy is 
the result of substituting y for w in fx.dx, dy is a single-valued 


Fb Fe Fb 
function of y, and f dy.dy = { dy.dy + f dy.dy, and there is 
Fa Fa Fe 


Fb b 
no error in using § ¢y.dy for f fx.dz. Butif x is a muitiple- 
Fa ig 


valued function of y, it will always happen that when y passes 
through a maximum, we pass from one set of values of x to 
another, and therefore from one set of values of dy to another, 
and in that aes ah is necessary to express our required integral 


as f¢ py.dy + co dy, taking pains to select the correct set of 
Fa 


values for py “ each integral. 

If y is a minimum between x=a and x«=B, essentially the 
same reasoning holds good. 

A couple of examples will make this clearer. 





2a 2 

(a) Take ede 

0/2 ax — a 
Let y=2ax—2°. Then ot = 2(a—x)=0 when x=a. 

da 
a ; 
~=—2,and y is a maximum when x= a. 
oe 





e=atvVa—y, 





dy 
day = = ——_*—.. 
2J/q?—y 
Since a is positive from «=0 to x=a, and negative from 
a 
t=O eT 2a, dxr= ia OY oh and w=a—va*—y from 
2Va?—y 
dy 


a=0 to «=a, and dve=— , and x=a+vVa°—y 





2Vati—y 
from x=a to «=2a. 


92 INTEGRAL CALCULUS. [ART. 88. 





Hence 
{- i = (“= xd a Re 
0V2ax—a MV2ax—x sat tas 
=sf- ‘a—Vae—y rane at+ve—y dy 
Vaty—y y? 
=if ‘a—Vae—y dae oe ate .dy 
Vary — y? 2 Vary —¥ 
a f, BE cak ST (Ex. 7, Art. 84) 
Wary — ¥? 
b fo 
(2) 0(sinw+ cosx) 


Let y = sinz + cosa. “ — cosa — sina = 0 when y=t 
xv 


2 ay 
Ia sinew —cosa#=—~V2 when «=. ~~ Therefore y has 
da? 4 


. > 7 
a maximum value V2 when «= mh 


main ), : 





any cos : GG ice oh BY 


5 V2 — 4? 





Since ot =Q and v4 <0 when «= it it follows that a is 


positive from «=0 to «= . and negative from «= r = to v= 4 
Hence we have 


2 4 ry 
I a 
hare + cos#)? Fee +cosxv) J,(sinx + cos)? 
4 


— ~v2__dy Wel te dy =2" da 
Space Jaro, Joes 


Cnap. VIII.] DEFINITE INTEGRALS. 93 


Let es sin 6; 
/2 


v2 
if ta All ai ll an esc?édd = ©, 
1 ¥YV2—y 27 2’ 


4 
and 


Si came cosy 
o(sing + cosa)? — 


EXAMPLE. 





(1) Show that ' (athe sees 2) a 


89. Differentiation of a definite integral. 

We have seen in Art. 51 that a definite integral is a function 
of the limits of integr ation, and not of the variable with respect 
to which we integrate ; that is, that fe fu.dx is a function of a 
and b, and not a function of x. Strictly speaking, “fede is 


a function of a and 3, and of any constants that fx may con- 
tain, where by constant we mean any quantity that is indepen- 
dent of a. 

If the limits a and 6 are variables, they are always indepen- 
dent of the x with respect to which the integration is performed, 
which must from the nature of the case disappear when the 
definite integral is formed, as it always may be in theory, from 
the indefinite integral; and this assertion holds good even when 
the same letter which is used for the variable with respect to 
which the integration is performed appears explicitly in the 
limits of integration. 


Thus if we write if, sing.da, the # in sinw.dx and the « which 
0 


is the upper limit of integration do not represent the same 
variable, and are entirely unconnected. Indeed, the former x 


94 INTEGRAL CALCULUS. [ART. 89, 


may be replaced by any other letter without affecting the value 
of the integral. For 


zx x 
f sin v.dx ={ sin z.dz = 1 — cosa. 
0 0 


Let us now consider the possibility of differentiating a definite 
integral. 


b 
Required D, f S(#,.a) dx, where a is independent of #, and 


a and 6 do not depend upon a. 





We have 
; we *,a+ Aa) dx — a.) Ce 
Df f(a, a)de = aint | £2 ae f fe ) i 
limit > f(%, a+ Aa) — 5a 
=e of ett IE) ap 
=f limit PA a + Aa) — f(a, »)) ae, 
ak Aa =i) Aa 
Hence D.{ F(a, a) dx SHRI: a) | dz, [1] 


and we find that we have merely to differentiate under the sign 
of integration. 

The truth of the converse of the last proposition can be eapele 
established, and we have 


Hf j if a, a) ae | da = f i f(a, a) de | da, 2) 


or even 


fi | f f(@, a) aa | da. =f) uf; f(a, «) de | da, [3] 


if a, b, c, and d are entirely independent. 


Suppose now that we are dealing with variable limits of 
integration. 4, 


Cuap. VIII.] DEFINITE INTEGRALS. 95 
Let us find first = f “fede. 
eda 


Let f fe. dx = Fx, then { JSu.da = Fz — Fa; and since by 


definition ade = fx, it follows that a aes 
dx dz 
Hence = { fe dn = rl =) = = fz. [4] 
In the same way it may be shown that 
b 
& { fede = — fz. [5 | 


Let us now take the most complicated case, namely, to find 


b 
= (fe, a) dx, where a and 6 are functions of a. 
Ae/a j 


Let if Ao e y JO RE a 
b 
then wu = { s@. a) da = F'(b, «) — F(a, a), 
ae du _dF(b,a) _ dF (a, a) ; 
dla. do. dla. 


but as 6 and a are functions of a, 


(etna) a) = D,F(b, a) +D, F(b, a), 
da 


and PER BERG! 0) OD, Fa, a), 
dla dla 
by I. Art. 200. 
D, F (6, «)=f(0, a), 
De ade 102 a): 


Hence au _ p, [F(b, «) — F(a, a)] +f (8, oS — Esta, 2) ee 


or 


£ (We: 4) dx = { (D.M@, a)) “apt Ce a) © db ena ; a) e 6] 


on 


96 INTEGRAL CALCULUS. (ART. 91. 
EXAMPLES. 
(1) = fin (x+y) dw =(x+1) sin (ay +y)—siny. 
Y 0 
Vz 
d ” 1 
2 SS Yt = = 
Se ah pian 
a ieee cle —————- 
(3) dd ) V1 — cosd.dd = e® V1 — cose?. 


90. When the indefinite integral cannot be found, the prob- 
lem of obtaining the value of the definite integral usually be- 
comes a more or less difficult mathematical puzzle, which can 
be solved, if solved at all, only by the exercise of great inge- 
nuity. Some of the results arrived at, however, are so impor- 
tant, and some of the devices employed so interesting, that we 
shall present them briefly here. But we must repeat the warning 
that most of the methods are valid only in case the definite 
integral is finite and determinate ; and erroneous results have 
more than once been obtained and published when a little atten- 
tion to the precautions described in Articles 83-86 would have 
prevented the mistake. 


91. Integration by development in series. 











1 , 
(a) f loge ae. (v. Art. 84, Ex. 1.) 
0 1l—2 
t =(l—at=l+e+et+at-.., fel. 
— wa 
1 . 1 
af; D8" da = { (loge +x loge +2? loge +--+) de, 
0 1l—@ 0 
1 
1 
"lor z.dz= — ——_—___.. . Art. 55 (a). 
fe og x.da (n+ 1) (v. At (a).) 
Therefore 
“1 log a bP es Riad Wig 8 T 
doe == — | A ee 
J Tesi! Garr ot ) § 


(v. Todhunter’s Trigonometry, Chap. XXIII., Ex. i.) 


Cuap. VIII.] DEFINITE INTEGRALS. 97 


(b) Ef} log (Sy) (v. Art. 86, Ex. 2.) 
0 aa 


log ce 7 = log fi = a) clipe yim LN de 























(I. Art. 130.) 
Hence 
ae x ad —8x —dx —7x 
log Pacts d= 2 { jet i al da 
0 e*—1 0 3 o 7 
1 1 1 
PRS Pay ie lhe cca 8 Gp seit lp 
( +atatat ) 
Teeth 1 1 1 
But gts ha il AN Copa oh ae gna all 
, gee 8 
(v. Todhunter’s Trig., Chap. XXIII., Ex. 1.) 
Therefore Jog (E Seton Vf tO, 
0 —l 4 
EXAMPLES 
(1) log? gy cama 
ol+2 12 
log a 3 
.d =— — 
(2) o1—2 8 


aa 
2 : 
‘leu 3) #+ * if <1, 


Ou, INTEGRAL CALCULUS. [ART. 92. 


Ray gL OD 1\2k2 /1.8\2i4 
@) J Ss aaa i () 1 (a) 3 
F\ 2 7.6 
ioe) aed if <1. 


92. Integration by ingenious devices. 





(a) log sin x.da. (v. Art. 84, Ex. 6.) 
0 
z 
Let u ={ log sinaw.da. 
0 
Substitute y= — oe. 


2 


0 2 
as —f log cos y.dy ait log cosa.da. 
7 0 


7 
“2 
2u = (log sin # + log cosa) dx ={ log (sin & cos x) da 
0 0 


2 arene 
= J log a ax 
0 2 


=— , log (2) + F log sin 2 a.da 


tol 9 


=— log (2)+ z { log sin a.da. 
2 2/0 
all log sinw.dx = { *log sin v.dx +f log sin w.dx 
0 0 7 
2 
=U +f log sinw.da. 
yi 


Substitute Y=r—4, 


Cnap. VIII.) DEFINITE INTEGRALS. 99 


and 
2 
: log sinw.dx = — "log siny.dy = log sinw.dx =u 
7 0 
Hence ‘aan 
and 
u= fi log sin w.dx =f log GOR te He dant log (2). [1] 
0 0 y ji 
(b) f, ede. (v. Art. 86 (a@).) 
0 
Let U =/! edz, and let r=ay; 
0 
then u =f ae dy =f ae" a, 
0 6 


ae 
we—e ={ oer Te da, 
0 


uf e-* da. = u" = 4K Gt rT ea Bal da) dx, by [3], Art. 89. 
0 0 0 





But 
(aerate dai= bs l 
0 2k + x 
If ode 
Hence w= a or = rt 
and f ePdx= s Vr. [2] 
0 2 
(c) 7 f Se (y. Art. 86 (b).) 
0 x 


fe] 


We have r=( e~tda if >0. (Art. 82, Ex. 6.) 


0 


100 INTEGRAL CALCULUS. [ART. 92. 


Hence 


“sin ma 
ii VESPA AL “ da = 
0 x i 


aif 


ak 


io 2) 


0 
Therefore 


en ® 
SID mx 
J TERESA OY 9 fo mee 
0 x 


— 
—— 


(1) {= log sina. da 


2) ff lox (#+ haces 


(3) feeeae 
(4) — 
Vlog — 


os) 0 : 
(5) f sin 2 Cosme Oy 
0 ay 





(sin ma f "da da 
0 
(J, e~°*sin mada) da 
0 


ah e-* sin mx. ae) da, by [3], Art. 89. 





nid (Art. 82, Ex. 7.) 
a 2m? 
7 
- if 0 
9 i m> 
™ if m<0 f? Ea 
2 
OWT Sais 0.0m by Art. 82, Ex. 5. 
EXAMPLES. 
2 
T 
= — — log PAW 
5 log (2) 
5=7 log(2). 
Suggestion: let «= tand. 
ke 
=3; if 
=Vr. 


=0 if m<—1 or m>l 
if m=—1 or m=l1 


if —Leam <1: 


Cuap. VIII.] DEFINITE INTEGRALS. 101 


T 


rene 
6) [SF deat 
cP) i a - 2 


93. Differentiation or integration with respect to a quantity 
(v. Art. 89.) 


Suggestion: integrate by parts. 


which is independent of x. 


(a) We have ff e-* da =i. 
0 a 


Differentiate both members with respect to a, 
vet aaah RO ie ae 
ane 


(Art. 82, Ex. 6.) 


°F baa 1 
— xe" dx) = — — ‘ 
of (—x a) 2 OY f, 


Differentiate again, 
: ! 
f eo dee = el 
0 a 


Differentiating » times, 


Page et, n! 
f ere de = aah [1] 
Va, (Art. 92, Ex. 3.) 


(6) We have f e-™ da = i 
0 2 a 


Differentiating n times with respect to a, 


(v. Art. 86, Ex. 3.) 











ere Ne Peso ee (20 — 1). be 
f, Cte Os — Sah a a [2] 
(v. Art. 86, Ex. 2.) 
: (Art. 82, Ex. 6.) 


(c) We have J ed a a 
0 
Multiply by dc, and integrate from «a to b, 
© b b 
f (J de) da ={ ue 
0 a a © 
© ,—az_ p— bu 
Hence f Pee heii log 2. [3] 
0 a a 


102 INTEGRAL CALCULUS. TART. 94. 


d “ad : 
oe” dae = . 
( ) i a+1 
Multiply by da, and integrate from b to a, 
1 a a 
it (f toe da. dx = da . 
0 b batl 


fel WG __. md 
Hence J se ~~ dex = log (3) [4] 











EXAMPLES. 


(1) From {sz == sah obtain 
oo ra 2 ~/q 





sf, 27 135.0(Qn—1), 1 
(a" oe 2 2.4.05 008 Jeni 


1 
(2) From J, de= obtain 
0 n+1 


arr m m m } 
{2 (logayrde = (— 1)" ae 


obtain 





: os) 
= a 
(8) From f e~—* cosmau.da = — ; 
0 a +m 


rf et e— b2 m 
{ —_—_—§— cosmw.dx = $log ( i 5} 
0 x a +m 


8) 
; m 
(4) From if e-* sinma.dx = -——._ obtain 
0 a“ +m 








nn —bx 
e~—e : a 
afi _ sin ma.dxa = tan7!— tan!— 
0 x m m 


94. The method illustrated in Art. 93 can be applied to 
much more complicated forms. 


(a) tf eB. dy, (v. Art. 86, Ex. 2.) 


Cuar. VIII.] DEFINITE INTEGRALS. 103 








Let “= one dx ; 
0 
then ba east f, tad -e° 
da ae 
Substitute Z= = 
du a ate Tho Sp ME ade 
and au of eo” #f#dz=—24 e *#dr=—2u. 
da 0 0 
Hence eee 2da. 
u 
Integrating, 
logu=—2a+C, 
and u=Ce**. 
When a=0, u =|) e@da=tvr. (Art. 92 (b)[2]-) 
0 
Therefore C,=4Nv7, 
ora) <= —2a = 
and u = if e” #da=° sc [1] 
0 2 
(dD) f e-™ coshba.da. (v. Art. 86, Ex. 2.) 
0 
Let u =|) e-”* cos ba.dx, 
du 8 + g2e? 2 
then —_—=— xe sin ba.du. 
db 0 


Integrating by parts, 


an 


ee epee me 
ne-@” sinbu.da =. | e *” cosbr.dx= 54 U. 
; 2 
2 a*J0 2a 


0 





du b 
Therefore FT ree. Us 
or | LU ec ea db. 


104 INTEGRAL CALCULUS. [ART. 95. 


or 


an 


Integrating, we have 





D2 
U = C; é a2, 


When 0=0, u ={ Biles br ma Zz, (Art. 92, Ex. 3.) 
0 Za 








ao a b2 
Hence U ={ e “ cos ba.da = Vo 4a2, [2] 
0 2a 
EXAMPLES. 
. ~e-*sin mx m 
(1) f, iam ere 9 {Lime eH ge oe 
0 x a 
(2) uN COSME | 1, ey 
1+27 Z 
1 toa) 
Suggestion : = 2 { ae Par) da. 
1+ 2? J0 


95. Introduction of imaginary constants. 


uf cos (x2) dx. (v. Art. 86 (c).) 
We have f Ban ce - Vr, (Art. 92, Ex. 3.) 
0 yam) 
Let = EV 1 = (cos +V—Isin5 } 
Then a=e(cost +V—I int) = a (1+ “/ 
(Art. 25.) 
d ia : (1 — Nae 


2a ~ eva (1 +V¥—1) eee 


Cuap. VIII.] DEFINITE INTEGRALS. 105 


Hence if) en Pe Vl day = age Se tea Hie 
0 2¢ N2 


But ee v-1 = cos (c?a*) — V—1 sin (c* 2"). 
({5] Art. 31.) 
Therefore 


{ cos (¢?a") da —V — rf Birt (Osa ete = Ne (Vninen 
a 0 0 ZC . 





and | 
nw io.) : 1 rT 
cos (ca) di =f s va a= isi 1 
f, S (c° a) dx ; in (¢c a") dx ND [1] 
(Art. 17.) 
Let ‘hh hi 
and if: cos (a) dx =( sin (a) dx = iE [2] 
0 0 2 . 
If we substitute y= 2’ in [2], we get 
“COSY *siny + 
~dy = —— dy =\/7 3 
Oa Y 0 VY 2 i?) 


Gamma Functions. 


1 n 
96. It was shown in Art. 84 that af (102 :) dx is finite and 
0 wx 


determinate for all values of n greater than —1, and infinite 
when n is equal to or less than —1. The substitution of 
y = log+ pees tue integral to f “yre-" dy, or, what is the 
same thing, to Mh we *dx; and in Art. 86, Ex. 3, the student 
has been required to show that this integral is finite and deter- 


minate for all values of greater than —1. 


a fp aa vi 7 —-1,- > 
fe “de =— a"e pn fa e-* da, 


by integration by parts. 


106 INTEGRAL CALCULUS. [ART. 96. 


If n is greater than zero, 
Te Ovum nena ss), 


and #"e-* is indeterminate when x=. Its true value when 
x= 0, obtained by the method of I. Art. 141, is, however, zero. 


Therefore f ee de = nf ote. 0x [1] 
0 0 
for all positive values of n. 


If n is an integer, a repeated use of [1] gives 


a 12) ao 
J rere ie =) if 2 un. 
0 0 


but f CEO hana 
0 


fo a) 
and we have f eerie ans [2] 
0 
provided that n is a positive whole number. 


If n is not a positive integer, but is greater than —l, 
—_o 


J v"e*dxz is a finite and determinate function of n, and its 
0 
value can be computed to any required degree of accuracy by 


methods which we have not space to consider here. 


ao 


fe tetade is generally represented by I'(n), and has been 


0 
very carefully studied under the name of the Gamma Function. 


If n is a positive integer, we have from [2] 


T(n+1)=n!. [3] 
From [3], T(2) coil [4 | 
Since Day =| Wenge =| edn, 

Ply Sols [5] 
We have always from [1] 

T'(n+1)=nIP(n), [6] 


if n is greater than zero. 


Cuar. VIII.] DEFINITE INTEGRALS. 107 


ao 
Since f x"e~*dz is infinite when » is equal to or less than 
0 


—1, it follows from the definition of T'(n) that I'(n)=o if 
m is equal to or less than zero. It has, however, been found 
convenient to adopt formula [6] as the definition of ['(n) when 
m is equal to or less than zero, and to restrict the original defi- 
nition to positive values of n. The result easily deduced is 
that I'(n) is infinite when n is equal to zero or to a negative 
integer, but is finite and determinate for all other values of n. 


97. We may regard the formula 
T(n+1)=nI(n) 


as a sort of reduction formula; and since each time we apply 
it we can raise or lower the value of n by unity, we can obtain 
any required Gamma Function by the aid of a table containing 
the values of I'(n) corresponding to the values of n between 
any two arbitrarily chosen consecutive whole numbers. 

Such tables have been computed, and we give one here con- 
taining the common logarithms of the values of I'(n) from 
m=1 to n=2. The table is carried out to four decimal 
places, and each logarithm is printed with the characteristic 9, 
which, of course, is ten units too large, the true characteristic 
being —1. 

10+ log T(m). 


n 0 1 2 3 4 5 6 7 Shi 




















_* 9975 |.9951 | 9928 | 9905 | 9883 | 9862 | 9841 | 9821 | 9802 
9.9783 | 9765 | 9748 | 9731 | 9715 | 9699 | 9684 | 9669 | 9655 | 9642 
9.9629. | 9617 | 9605 | 9594 | 9583 | 9573 | 9564 | 9554 | 9546 | 9538 
9.9530 | 9523 | 9516 | 9510 | 9505 | 9500 | 9495 | 9491 | 9487 | 9483 
9.9481 | 9478 | 9476 | 9475 | 9473 | 9473 | 9472 | 9473 | 9473 | 9474 











9506 
9.9511 | 9517 | 9523 | 9529 | 9536 | 9543 | 9550 | 9558 | 9566 | 9575 
9.9584 | 9593 | 9603 | 9613 | 9623 | 9633 | 9644 | 9656 | 9667 | 9679 
9.9691 | 9704 | 9717 | 9730 | 9743 | 9757 | 9771 | 9786 | 9800 | 9815 
9.9831 | 9846 | 9862 | 9878 | 9895 | 9912 | 9929 | 9946 | 9964 | 9982 


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108 INTEGRAL CALCULUS. [ART. 97. 


Such a table enables us to compute with Gamma Functions 
as readily as with Trigonometric Functions, and consequently 
the problem of obtaining the value of a definite integral is 
practically solved if the integral in question can be expressed 
in terms of Gamma Functions. 

For example, let us consider 


eo 
(a) Hf) ee lan. 
0 


Let Of = (hs 
then ees da —_ af e7y dy — ree ee da. 
0 "yeas arto 
Hence f ee “dae = i, [1] 
provided that a is positive and n>—1. 
1 1 n a 
(b) if on (tog *) dx. (v. Art. 84, Ex. 4.) 
0 x 
Let , y =— loga. 
1 n 9) 
then f ri (10g 5) dx ={ y” e (m+l1)y dy. 
a/0 4) 0 


Henee, by [1], 


abit 1\* T (n 4-1) 
ais k = C= 2 
J, : ( °8 ;) ee (m+ Pyeey bad 


if m>—1 and n>—l1. 


then f, etde=t —dy=4f a te-* da. 
0 0 /y 0 


Hence f e-* dw =4T (4). [3] 
0 


Cuap. VIII.] DEFINITE INTEGRALS. 109 


a 1 
98. iff x” (1—2)" de = B(m, n) [1] 
0) 


is an exceedingly important integral that can be expressed in 
terms of Gamma Functions; it is known as the Beta Function, 
or the First Eulerian Integral, '(n) being sometimes called the 
Second Eulerian Integral. 

In the Beta Function, m and n are positive, and B(m,n) is 


always finite and determinate. . (v. Art. 84, Ex. 5.) 
1 
In ea (1—«a)"""dxw let y=1—a, 
ef) 
and we get 


1 wy 
fh gmt (1 oe Wate da =| ater % (1 ee are dy, 
0 0 


or B(m,n) = B(n, m). [2] 





1 
In ft a—ay as let eae, 
0 


and we get 


[2 ml (1 x8 oe Ge aac a a 
fey ae 


ce m—1 

He J Rane iy a By ye 3 
nce > Gta (m, 2) [3] 

We have seen in [1] Art. 97 (a) that 

{ Nm pax Ly (n aE Ly: 

Ef ae” da Soir pecrs ry 
Hence T'(m) =f Grae ere dx, 
0 


oo 
T'(m) qn} e-4 =i) qntn-1 gl e~a+2) dx, 
0 


P(m) {- a" lo -—4*dq ={2 gm (S @ qntn— —le earnda) dx, 


T(m) T(n) = f “yn ara ® 


110 INTEGRAL CALCULUS. [ARrT. 99. 


Therefore Ey ie ae ={ dx 5 [4] 
T(m+n) (1 reas 
or by [3], 


a if »m—1 __ rn—l ral T(m) I'(n) A 
B(m, n) ={ a1 (1 — 2) Ot | 5 aaa [5] 


If n=1—~mM, then since [T(1)=1, 
ie ee - de = P(m) PA + 6 
Formula [6] leads to an interesting confirmation of Art. 


92 (b). 
Let m= 4, and we have from [6] 


da 
(rd) } =U, vacua 


Substitute CRANES 





[e<) 


0 +a) Vz 0 1+y¥ 


and we have 





Hence | Ia) so [7] 
and since by Art. 97 (c) 


{i etde  =4TQ), 
0 

ioe] 
fi etde | =bye. 
0 


99. By the aid of formulas [4], [5], and [7] of Art. 98 
a number of important integrals can be obtained. 
For example, let us consider 


Tv 


2 e e 
if sin" 2.da, where n is greater than — 1. 
0 


Let y= sing, 


yal 


> 1 
and we have f sin” @.dx =f) y" (1 — yy’)? dy. 
0 0 


Cuap. VIII.] DEFINITE INTEGRALS. lil 


Let, now, pan NE 
and 
1 6 1 1 ae 1 
f y" (1 —y’) dy = af (1 —2)72dz 
0 0 


a] n+l 


=$ (2? ‘ah de=4B(755, 5) 
0 2 2 


nd 

















But 
1 
r(“$*)r@) 
B an Ei eeetweeect Aiicie: by [5] Art. 98. 
Bast 2 G 
hapa 
2 
r « + } 
ol 5 a NCO by [7] Art. 98. 





Hence 
ih sin” «2.da = Va rst a 
0 2 7 


[1] 


If n is a whole number, this will reduce to the result given 
in Art. 87, Ex. 5. 








EXAMPLES. 
a) wo” dat ein + 4) 
ae TOU + 1): 








r m+i1 r n+1 
2 2 


2 
(2) sin" x cos” x.dx = 
if eo 1) 





did INTEGRAL CALCULUS. 
: a i 
(3) dx es va n 
0 J] tas ot nr Tr 1 i 
Te oe 


l(p+1) r(=*") 
n 


el] 
4 m(] —y"\Pdx— - 
(4) fon eee "t*) 


PU aMiarc 


[ART. 99. 


Cuap. IX.] LENGTHS OF CURVES. 113 


CHAPTER DX. 


LENGTHS OF CURVES. 


100. If we use rectangular codrdinates, we have seen (I. Art. 
27) that 


dy 

tan aa [1] 
and (I. Arts. 52 and 181) that 

ds? = da” + dy’. [2] 

f dy 
From these we get sint = 70 [3] 
cos = a [4] 

ds 


by the aid of a little elementary Trigonometry. 

These formulas are of great importance in dealing with all 
properties of curves that concern in any way the lengths of arcs. 

We have already considered the use of [2] in the first volume 
of the Calculus, and we have worked several examples by its 
aid in rectification of curves. Before going on to more of the 
same sort we shall find it worth while to obtain the equations of 
two very interesting transcendental curves, the catenary and the 
tractrix. 


The Catenary. 


101. The common catenary is the curve in which a uniform 
heavy flexible string hangs when its ends are supported. 

As the string is flexible, the only force exerted by one portion 
of the string on an adjacent portion is a pull along the string, 
which we shall call the tension of the string, and shall represent 
by 7. T of course has different values at different points of the 
string, and is some function of the codrdinates of the point in 
question. 


114 INTEGRAL CALCULUS. [Arr. 101. 


The tension at any point has to support the weight of the por- 
tion of the string below the point, and a certain amount of side 
pull, due to the fact that the string would hang vertically were 
it not that its ends are forcibly held apart. . 

Let the origin be taken at the lowest point of the curve, and 
suppose the string fastened 
at that point. 

Let s be the are OP, 
P being any point of the 
string. As the string is uni- 
form, the weight of OP is 
proportional to its length ; 
'-——X_ we shall call this weight ms. 

This weight acts verti- 
cally downward, and must be balanced by the vertical effect of 7, 
which, by I. Art. 112, is 7'sinr. 

Hence Tsint = ms. (1) 


Y 





As there is no eaternal horizontal force acting, the horizontal 
effect of the tension at one end of any portion of the string must 
be the same as the horizontal effect at the other end. In other 


words, T cost =C (2) 


where c is a constant. Dividing (1) by (2) we get 


Cc 
$= —tanr, 
m 
or s=atanr, (3) 
where a is some constant. From this we want to get an equa- 
tion in terms of # and y. 


Ad 21 Vet ek Sk 2 
tanr =Vsec?s —1= ——13 
(Ls? 
h e-= 2 adh Lad 1 5 
ence a a ) 
or a? ds? = (a? + s*) dz’, 


ads ope Integrate both members. 


and Serre ye tee ae 
(a? + s*)3 | 


Cuap. IX.] LENGTHS OF CURVES. 115 


alog(s+Ve+s)=xe2+C; 


when x= 0, s=0, 


hence C=aloga, 
and log(s + Va?+ s*)=— + loga, 


Leaner ni z 
s+Ve+s =aes, 
—— z 
Ve+s? =aea—s, 
ee x 
a? = a’e*?a — 2aecs, 


ae 5 (ea — e7a) =<aetanr, “by, (3). 


Hence een pear ay: 
des: 2 
and y=" (c¢ Lema) 4 6) 


If we change our axes, taking the origin at a point a units 
below the lowest point of the curve, y=a when x=0, and 
therefore C= 0, and we get, as the equation of the catenary, 


y= SF (er+ ere). (4) 


EXAMPLE. 


Find the curve in which the cables of a suspension-bridge 
must hang. Ans. A parabola. 


The Tractria. 


102. If two particles are attached to a string, and rest on a 
rough horizontal plane, and one, starting with the string stretched, 
moves in a straight line at right angles with the initial position 
of the string, dragging the other particle after it, the path of the 
second particle is called the tractriz. 

Take as the axis of X the path of the first particle, and as 
the axis of Y the initial position of the string, and let a be 


116 INTEGRAL CALCULUS. (Arr. 102, 


the length of the string. From the nature of the curve the 
string is always a tangent, and we shall have for any point P 


y ; 
=~ Sint, [1] 


for 7 lying in the fourth quadrant has a negative sine. 





a el ei Blais NL) 
Maat ca T= a ~ da? + dy’ 
hence y? (da? + dy’) = a? dy’, 
yf dat = (a" — y*) dy’, 
a? —y’)id 
and dx aso 


is the differential equation of the tractria. 


On the right-hand half of the curve 7 is in the fourth quadrant, 
” or tan7z is negative, and we shall write the equation 
a 
a’ — y’)id , 
dx = — ma hike ; yey, [2] 


If we allow the radical to be ambiguous in sign we shall get 
also the curve that would be described if the first particle went 
to the left instead of to the right. The tractrix curve, generally 
considered, includes these two portions. 

Integrating both members of [2], and determining the arbi- 
trary constant, we get 


v= VER H+ alogt VON Y 


[3] 


as the equation of the tractrix. 


Cuap. IX.] LENGTHS OF CURVES. 117 


EXAMPLES. 


(1) Show by Art. 102 (1) that in the tractrix s=a log” 
if s is measured from the starting-point. J 


(2) Find the evolute of the tractrix. (I. Art. 95.) 


Rectification of Curves. 


103. In finding the length of an arc of a given curve we 
can regard it as the limit of the sum of the differentials of the 
arc, and express it by a definite integral. 


r= 


We shall have s= = fv Vda? + dy’. 


Of course in using this formula we must express Vda? + dy? 
in terms of « only, or of y only, or of some single variable on 
which « and y depend, before we can integrate. 

For example ; let us find the length of an arc of the circle 


of = a. 
2u.dx + 2y.dy =0, 


a.da 








dy = — ; 
y 
de? + dy? = EY ae? = © ae? = —* aa? 
a fat ye ) 
ee eo iat a(sin-2 sin") 
a ~/q? — a? a 
The length of a quadrant = a 2 ae a = 
\Teoe _ 2 


*, the length of a circumference = 2 7a. 


118 INTEGRAL CALCULUS. [ArT. 104. 


Length of Arc of Cycloid. 
104. For the cycloid we have 





ay seem res 
y=a —acosé c. “ae 
dz = a(1 — cos6) dé = ydé, 
6 = vers”, 
a 
lO pod as a de 
a Note V2Qay—¥° 
Cate 
tie ydy 5 
V2 ay — 7? 
ds? = da? + dy = 2 aydy” tn 2 ady" ; 
2ay—y 2a—y 
ds = V2a.—4_, 
2a— y 
CeO er ~~ oa 
s= J/2a —_“¥ _ — 2/2a(V2a— y —V2a—4). 
Y% V2a—y 
If the arc is measured from the cusp, % = 0, 
s=4a—2V2avV2a—%. [1] 
If the arc is measured to the highest point, y, = 2a, 
s=4a% 
The whole arch = 8a. 
EXAMPLE. 


Taking the origin at the vertex, and taking the direction down- 
ward as the positive direction for y, the equations become 
x=ab+asind ; 


y= a—acosé (I. Art. 100.) 


Show that s=2~/2ay when the arc is measured from the 
summit of the curve. 


Cuar. 1X.] LENGTHS OF CURVES. 119 


105. We can rectify the cycloid without eliminating 0. 
x= ad — asind 
y= a—acosé | 
dx = a(1 — cos) dé, 
dy = asin6.dé, 
da? + dy? = 2.a°dé?(1 — cos6), 


0, 
and s=ay2f (1+ cosd)'a6, 
9p 
9 : 5 9, 
et v2 fe 2 ee dé = ta {sin eas = 4a (cos “ _ cos} 
If 6:=0 and 6, =27, we get s = 8a as the whole curve. 


106. Let us find the length of an arch of the epicycloid. 


*=(a+b)cosé—b cos (t+ 4) 4 
, (1. Art.109[1].) 


y=(at+b) sin@ —b sin + "¢ 


dx =|- (a+b) sn6+(a+b) sin? “6 Jao, 





dy =| (a+b) cos6é—(a+b) cos + “6 |e. 





ds? =(a+ by a6 2— 2 (cost "4 cos + sin= =e b9 sind) 
=2(a+ byeaer(1 ~~ cos). 
9, a .\3 
s=(a+d)V/2 [ (1- “i dd, 
9% 


= REED cos 8 — cos — aif i [1] 


To get a complete arch we must let 6)=0 and PAPAL ia 
Cb 


Hence, for a whole arch, 
_ 8b(a +b) 


Gh 


120 INTEGRAL CALCULUS. (ArT. 107. 


EXAMPLES. 
(1) Find the length of an arch of a hypocycloid. . 
8b(a—b) 
ANn8, See ae 


(2) Find the length of an arch of the curve a +y =a’, 
and show that it agrees with the result of Ex. 1. 
(v. I. Art. 109, Ex?2,) 


107. Let us attempt.to find the length of an arc of the ellipse 











on ¥ 
amar b? = ‘le 
We have oO ees ee =(, 
a’ b? 
b?ax 
oe ee 
dy ay ) 
ga OOF 
bia? + aty? Pawo apa e a? — e? a? 
dal tT Oe Gye a da? = SE gaye 


aty? a? — x ve—x 


where e is the eccentricity of the ellipse. 


i xy O = 
Hence o=| fae al dit. [1] 


The length of the elliptic quadrant is 


oe , [2] 


These integrals cannot be obtained directly, but 


a! — ear |e 14.01 —e*) a" |2 
Q? — 22 a2 — 2 
can be expanded by the Binomial Theorem, and the terms of 
the result can be integrated separately, and we shall have the 


required length expressed by a series. 
A more convenient way of ere with the problem is to use 


an auxiliary angle. Instead ofiee 2 i= 1 we can use the pair 
of equations ML Vote 
peetete } (I. Art. 150), 
y=bcos¢d 


Cuap. IX.] LENGTHS OF CURVES. 121 


dx=acos¢.dd, 
dy =— bsin¢d.dd, 
ds’ = (a? cos’ + b’sin’d)d¢? = [a? — (a? — b*) sin? p |d¢? 
2 2 
A) hd € ore — u sint@) do’ = a?(1— e’sin’¢) d¢’, 
a) 


where e is the eccentricity of the ellipse. 





s=a "a —e?sin?)*dd [3] 
0 
= fu —te’sin’d —4-te*sin‘d —4-4-3e’sin®h---]d¢. 
0 
For the are of a quadrant we have 
sy= a) ra —e*sin’¢]?d¢. [4] 


EXAMPLES. 


(1) Obtain s, as a series from [2], and also from [4], and 
compare the results with Art. 91, Ex. 5. 


(2) Show that the length of an arc of the hyberbola is 
oy 272 } 
s= Df [1+ a" sinh? $ | ad. 
bo b° 


Polar Formule. 
108. If we use polar codrdinates we have 
ds = Vdr? +7?d¢?, (I. Art. 207, Ex. 2.) 
rd 





tane = ae (I. Art. 207.) 
From these we get, by Trigonometry, 
; rdd dr 
2 a cose = —: 
ds ds 


109. Let us find the equation of the curve which crosses all 
its radii vectores at the same angle. Here 
rdd _ 


tane=a, a constant, 
adr 


9 


ad 
at — dd, 
i 


122 INTEGRAL CALCULUS. (Arr. 110. 


alogr=¢+C, r=e 
ine a @) 
where 0 is some constant depending upon the position of the origin. 


This curve is known as the Logarithmic or Equiangular Spiral. 


110. To rectify the Logarithmic Spiral. We have, from 


LOD (1), A ; 
— = log—; 

a b 

ad = any 

rdd= adr, 


ds? = dr? + 7? dd? = (1 + a?) dr? ; 
s= fC + a*)kdr = (1 +02)4(7, — 1%). 
To 


EXAMPLES. 


(1) Find the length of an arc of the parabola from its polar 
equation 


m 
t= cc“. 
1+cos@¢ 
(2) Find the length of an arc of the Spiral of Archimedes 
T= 00. 


111. To rectify the Cardioide. We have 
r= 2a(1—cos¢), (I. Art. 109, Ex, 1), 
dr = 2asing.dd, 
ds’ = 4a’sin’ d.d¢* + 407(1 — cos ¢)?d¢? 
= 8a?d¢*(1— cos¢), 


Bee aA eis afi —cos ¢):dp=8 a| cos $t—cos | 


= 16a for the whole perimeter. 


Cuap. IX.] LENGTHS OF CURVES. 123 


Involutes. 


112. If we can express the length of the arc of a given curve, 
measured from a fixed point, in terms of the codrdinates of its 
variable extremity, we can find the equation of the involute of 
the curve. 

We have found the equations of the evolute of y = fx in the 
form 


a’ = & — p COSY 
; : \. (I. Art. 91). 
Y¥=Yy—psinv 
We have proved that tany=tanr'’, (I. Art. 95), 
! 
and that = = (I. Art. 96) ; 
sin 7’ = - 
es (Art. 100). 
cosr’ = ed 
3! 
Since tany=tanr', v=r’ or v=180°+7'. 


As normal and radius of curvature have opposite directions, 
we shall consider v = 180°+7'. 


Then sinv=—sinr’ and cosy=—cosr'’. 
dae! 
H — eae) 1 
ence x cy Bl er (1) 
pe ay : 
DAT or (2) 
Since dp = ds’, 

p=s'4l (3) 


where / is an arbitrary constant. Since # and y are the codrdinates 
of any point of the involute, it is only necessary to eliminate 2’, 
y', and p by combining equations (1), (2), and (3) with the equa- 
tion of the evolute. 

As we are supposed to start with the equation of the evolute 
and work towards the equation of the involute, it will be more 
natural to accent the letters belonging to the latter curve instead 


124 INTEGRAL CALCULUS. (ArT. 112. 


of those going with the former; and our equations may be 
written 


peal tp; y= y+ o's; fae ic 


Since p'=/ when s=0, it follows that / is the free portion 
of the string with which we start. (I. Art.97.) By varying / 
we may get different involutes of the same curve. 

To test our method, let us find the involute of the curve 

TOA: (7 — my’, 
SOM eR BL) 
for which /=m. We must first find s. 


2 ydy = = (a —m)* da, 


ya 
9m y 
ds? —_— 2x +m bevels 
3m 





$= — (2x +m) ide = 2a+m)i—m, 
V3 Tad, iri 








p=stms 1 _(x4m)i, 
VIM 
‘day mud 
Qa ae 
a 3 
PR 8 sisi 5 4 (2Qa+m)(x— di) 
7m y 
1 &—mMm 
f=? 
3 
A npl2 
y= @—m)=— 2, 
dy y 
x= 3a'+m, 
4 o!? 
Yo 
y' 
Substituting in (7) the values of x and y just obtained, we have 
y= 2mx' 


as the equations of the required involute. 


Cuap. IX.] LENGTHS OF CURVES. 125 


EXAMPLE. 


Find an involute of ay’? = 2’. 


113. An involute of the cycloid is easily found. Take equa- 


tions I. Art. 100 (C). 
x= ab6+asind 
y=—a+tacosé 


Let pres. 


dx=a(1+cosé)d@ =2a vos*S db, 


Mg 
=— 2a sin’ cos& dé, 


dy =—asin6.d6 3 


ds? = 2a°d6’(1+ cos6) = 4a? dé? cost, 


: cea iad 
s=2a{ cos-dé =4asin-; 
0 2 2 
Pray Gipoet G Petes he 
e= © + 4asin>coss = x +2asiné, 


y=y'+4a sin’ = y' —2a(1—cos6), 


z'=«ad—asing 
y' = a—acosé 


a cycloid with its cusp at the summit of the given cycloid. 


EXAMPLE. 


From the equations of a circle 
x= acosd 
y=asind 
obtain the equations of the involute of the circle. Let /=0. 


Ans. #'=a(cos¢+¢sin eh 
y'=a(sind—dcosd) )- 


126 INTEGRAL CALCULUS. (ART. 114. 


Intrinsic Equation of a Curve. 


114. An equation connecting the length of the arc, measured 
from a fixed point of any curve to a variable point, with the 
angle between the tangent at the fixed point and the tangent at 
the variable point, is the intrinsic equation of the curve. If the 
fixed point is the origin and the fixed tangent the axis of X, the 
variables in the ¢ntrinsic equation are s and r. 

We have already such an equation for the catenary 


s=atanr, Art. 1013); 44 
the origin being the lowest point of the curve. 
The intrinsic equation of a circle is obviously 
Sex Or, | [2] 
whatever origin we may take. 


The intrinsic equation of the tractrix is easily obtained. We 
have 


y=— asin, Art. 102 (1), 
and Sm log@ ; Art. 102, Ex. 1. 
A 
hence s = alog(— eser) 


where 7 is measured from the axis of X, and s is measured from 
the point where the curve crosses the axis of Y. As the curve is 
tangent to the axis of Y, we must replace + by s — 90°, and we 
get 
s = alog seer [3] 


as the intrinsic equation of the tractrix. 


EXAMPLE. 


Show that the intrinsic equation of an inverted cycloid, when 


the vertex is origin, is 
s=4asinr; (1) 


when the cusp is origin, 1s 
s=4a(1—cosr). 3 (2) 


Cuap. IX.] LENGTHS OF CURVES. 127 


115. To find the intrinsic equation of the epicycloid we can 
use the results obtained in Art. 106. 








da=(a-+0)(sin5F9 — sind d= 2(a+b) cos — “6 sin 6. dé, 


42h ns; 6. dé, 








dy =(a +5)(cos6 —cos 0) 10= 9(a-4.b)sin 
by the formulas of Trigonometry ; 
sin a — sinB = 2cos$(a+) siné(a—£), 


cos B— cosa = 2 sin}(a + B) sin$(a— B£) ; 








tant = 7 tan“ tS "6, 
hence 3 r=4 aria ; 
= ca — c08 558) by Art. 106[1]; 
therefore $= ee ¢ Tian a 57, *) [1] 


is the intrinsic equation of the epicycloid, with the cusp as origin. 
If we take the origin at a vertex instead of at a cusp 


= 4o(a+ 0), a 
a 
_7(a+2b) + 20) 
2a 5 
»  4b(a+)) Cie ie 
hence Roos SEGAL sin Aeon: Saree 


or . 
a a+2b 


is the intrinsic equation of an epicycloid referred to a vertex. 


128 INTEGRAL CALCULUS. (ART. 116. 


EXAMPLE. 


Obtain the intrinsic equation of the hypocycloid in the forms 


eee — cos“ 57), (1) 





a 
4b(a—bd) .. a 


116. The intrinsic equation of the Logarithmic Spiral is found 
without difficulty. 


$ 
We have ery (Art. 109), 
and. s=V1+a2(7,—%). (Art. 110). 


If we measure the arc from the point where the spiral crosses 


the initial line, 7) = 0, and we have 
1) 


s=bdvI1 DP i ee ry 
In polar codrdinates r = ¢ + «, and in this case «= tana; if 
we measure our angle from the tangent at the beginning of the 
arc we must subtract « from the value just given, and we have 


5 b(NIa a (ECan 


or, more briefly, s=k(c’—1), k and c being constants. 


117. If we wish to get the intrinsic equation of a curve directly 
from the equation in rectangular coordinates, the following method 
will serve : 

Let the axis of X be tangent to the curve at the point we take 
as origin. 

dy , 
tan; = ae (1) 


and as the equation of the curve enables us to express y in terms 
of x, (1) will give us x in terms of 7, say 7 = F7; 


Cuap. IX.] LENGTHS OF CURVES. 129 


then dx = F'r.dr, divide by ds ; 
dx dr dx 
a + Rai! but —= . 
a AB u re COST 3 
hence ds = secrF"r.dr. (2) 


Integrating both members we shall have the required intrinsic 
equation. 


For example, let us take x? = 2my, which is tangent to the 
axis of X at the origin. 


2xdx = 2mdy, 


ue tant =—, 
dx m 


dx = m sec’r.dr, 








dx dr 
— = cost = msec’?r—, 
ds ds’ 
ds = msec’r.dr, (1) 
dr m| sinr iF 
s= = — logtan{- + - CO 
mf cos*r 2 ie ia Gi 5 5) hia, 
$=0 when r=0; o2 O= 0; 
m{ sintr T 
= —| —— + logt —}}. 2 
2 = i oe G rh 5) | oe 
EXAMPLES. 


(1) Devise a method when the curve is tangent to the axis 
of Y, and apply it to 7? = 2m. 


(2) Obtain the intrinsic equation of y’? = = (2 —m)?*. 
m 


(3) Obtain the intrinsic equation of the involute of a circle. 
(Art. 118, Ex.) 


130 INTEGRAL CALCULUS. _ (ART. 118. 


118. The evolute or the involute of a curve is easily found 
from its intrinsic equation. 





If the curvature of the given curve decreases as we pass along 
the curve, p increases, and 


s' = p— po. “CL Artes). 


If the curvature increases, p decreases, and 
p 


S! = po — p- 
Hence always s'=+(p— po); [1] 
ene (I. Arts. 86 and 90). 
dr 


We see from the figure that 7! =r. 


Hence treo Coe was me 2 ’ 
: dr T=7! dr T=0 


or, as we shall write it for brevity, 


sm [2] 


dr, 


119. The evolute of the tractriz s = a logsecr is 


eee a SeC7|' _atanr, the catenary. 
i 0 


The evolute of the circle s = ar is 


. 
Sez al =0, a point. 
dr|, 


Cuap. IX. ] LENGTHS OF CURVES. 131 


The evolute of the cycloid s = 4a(1— cos7r) is 


Nee 4g HU= cost)" 


=4asinr, 
dr 


0 


an equal cycloid, with its vertex at the origin. 


EXAMPLES. 
(1) Prove that the evolute of the logarithmic spiral is an 
equal logarithmic spiral. 
(2) Find the evolute of a parabola. 


(3) Find the evolute of the catenary. 


‘120. The evolute of an epicycloid is a similar epicycloid, with 
each vertex at a cusp of the given curve. é 
Take the equation 


pee nO Mai Oy fy Pode eee rt ATES TD Piss 
cl a+2b 
For the evolute, 


4b (a+b) © 


ca 


Be Ae ein ers [1] 
a+2b a+t2b 








The form of [1] is that of an epicycloid referred to a vertex 
as origin; let us find a’ and 0b’, the radii of the fixed and rolling 
circles. 


4b'(a'+0') .. a! eras 
s= cae 35) T; by Art. 115 [2]; 





a a'+ 20! 
! ! ! 
hence, pond 10 ) aa by SOO) ae i ; 
a’ a 


a+ 2b' a+2b 


132 INTEGRAL CALCULUS. (Arr. 121. 


Solving these equations, we get 








al= i 
at2bd’ 

es ab 
a+2b’ 

a’ _o 

ned 


and the radii of the fixed and rolling circles have the same ratio 
in the evolute as in the original epicycloid; therefore the two 
curves are similar. 


EXAMPLE. 


Show that the evolute of a hypocycloid is a similar hypo- 
cycloid. 


121. We have seen that in involute and evolute + has the same 
value; that is, r=7’. 

If s' and 7' refer to the evolute, and s and 7 to the involute, we 
have found that 


or s'=—-—l, 1 being a constant, 
the length of the radius of curvature at the origin. 
(s'+1)dr'=ds, 
s= (s+ 1)dr' 


is the equation of the involute. 
The involute of the catenary s = a tanr is, when /= 0, 


Ee 
s=a | tanr.dr=alogsecr, the tractria. 
0 


Cuap. IX.) LENGTHS OF CURVES. 183 
The involute of the cycloid s = 4asinr when /= 0 is 
ov 
a 4a sinr.dr = 4a(1—cosr), 
0 


an equal cycloid referred to its cusp as origin. 
The involute of a cycloid referred to its cusp s =4a(1 —cosr) 
when / = 0 is 


x 
s=4af (1—cosr)dr=4u(r +sin7), 
0 


a curve we have not studied. 
The involute of a circle s = ar when 1 = 0 is 


a ar? 
$=a| r.dr = —. 
0 2 


122. While any given curve has but one evolute, it has an 
infinite number of involutes, since the equation of the involute 


‘= {(s + 1)dr 


contains an arbitrary constant /; and the nature of the involute 
will in general be different for different values of J. 

If we form the involute of a given curve, taking a particular 
value for /, and form the involute of this involute, taking the same 
value of /, and so on indefinitely, the curves obtained will con- 
tinually approach the logarithmic spiral. 

Let $= jr (1) 
be the given curve. 


s=f (+ fart + [fear 


is the first involute ; 


T 2 T 
8 = |i + lr +f fr.dr) dr = Ir + = +f fr.dr? 


is the second involute ; 
Ir? Ir? lr” 4) pr 
s=lr+—+—4+ 0 a ah f St.dt (2) 


is the nth involute. 


134 INTEGRAL CALCULUS. TART. 123. 


By Maclaurin’s Theorem, 


=fot for Fst Sh svot “a 
But s=0 when r= 0; hence fo=0, and 
frm Air +27 hie seeee ; 
Chae atte Ae Ay 
Fear a ae vssele ; 


Dodi tes Antes 
r= ~ ane 3 
we ug ee Gea). (eee he? 
as n increases indefinitely all the terms of (3) approach zero 
(I. Art. 133), and the limiting form of (2) is 


i aire 
Sante ae bis 
Ts ra a 
= Cosesrray mp ti aD 
s=1(e7— 1) by I. Art. 133 [2], 


which is a logarithmic spiral. 


123. The equation of a curve in rectangular coordinates is 
readily obtained from the intrinsic equation. 


Given eee Ror 
. adn 

we know that sint = an 
ds 

dx, 

and cost = —;}; 
ds 

hence dx = costds = coszf't.dr, 


dy = sin rds = sin a dr, 
w= [ cosrs"r. dr L 


y= [sin ty ey dr | 


Cuap. IX.] LENGTHS OF CURVES. 135 


The elimination of + between these equations will give us the 
equation of the curve in terms of # and y. Let us apply this 
method to the catenary. 


s=atanr, 


ds = asec’r.dr, 


ei a { secr.dr = ¢ fed ee 
0 


Ps 7 
1—sinr 


: 
y= af seer tanr.dr = a(secr — 1), 
0 





e+ +] ea aa ena 
secr=4(er7+€e «), 
y= 5 (e+e) —a, 


the equation of the catenary referred to its lowest point as origin. 


Curves in Space. 


124, The length of the arc of a curve of double curvature is 
the limit of the sum of the chords of smaller arcs into which the 
given arc may be broken up, as the number of these smaller arcs 
is indefinitely increased. Let (a, y,z), (vw+dx, y+ Ay, z+ Az) 
be the codrdinates of the extremities of any one of the small arcs 
in question; dx, Ay, Az are infinitesimal; Vdz?+ Ay’?+ Az? is the 
length of the chord of the are. In dealing with the limit of the 
sum of these chords, any one may be replaced by a quantity dif- 
fering from it by infinitesimals of higher order than the first. 
Vda? + dy’? + dz? is such a value ; 


hence s= | Vde + dy? + dz. 


r= XH 


136 INTEGRAL CALCULUS. (Arr. 124. 


Let us rectify the helix. 
x=acosé | 


= 51 8 (1. Art. 214.) 
Cue, J 
da = — asin6.dé, 


dy = acos6.dé, 
dz = kdé, 
ds? = (a? +k”) d6’, 
6, Che dc. 
s= (a? + k*) if do = Vae+ k? (6, — ). 


¢ 


EXAMPLES. 


ad 
(1) Find the length of the curve {y= a pel : 
20 60 
Ans. S=a2+2+1. 
[ae 


(2) y= 2Vae— a, 2=%—FyR Ans. s=u+y—zZ. 


Cuap. X.] AREAS. 1387 


CHAPTER X. 


AREAS. 


125. We have found and used a formula for the area bounded 
by a given curve, the axis of X, and a pair of ordinates. 


A = { yde. 


We can readily get this formula as a definite integral. The 
area in the figure is the sum of the 
slices into which it is divided by the 
ordinates; if Aw, the base of each 
slice, is indefinitely decreased, the 
slice is infinitesimal. The area of 
any slice differs from yAa by less 
than AyAw, which is of the second 
order if Aw is the principal infini- 
tesimal. We have then 





UA ee eaten & ye by I. Art. 161. 
Hence A = yde. [1] 


If the curve in question lies above the axis of X, and % is 
less than 2, each ordinate is positive, each Aw is positive, each 
term of the sum whose limit is required is positive, the sum is 
positive, and the limit of the sum or the area sought is positive. 
If, however, the curve lies below the axis of X, and 4 is less 
than x,, each ordinate is negative, each Aw is positive, each 
term of the sum is negative, the sum is negative, and the limit 


138 INTEGRAL CALCULUS. [ART. 125. 


of the sum or the area sought is negative. If, then, the curve 
happens to cross the axis between 2 and 2, formula [1] gives 
us the difference between the portion of the area above the axis 
of X and the portion below the axis of X, but throws no light 
upon the magnitudes of the separate portions. Consequently, 
in any actual geometrical problem it is usually necessary to find 
the portion of the required area above the axis of X and the 
portion below the axis of X separately; and for this purpose 
it is essential to know at what points the curve crosses the axis. 
Indeed, if the problem is in the least complicated, it is neces- 
sary to begin by carefully tracing the given curve from its 
equation, and then to keep its form and position in mind during 
the whole process of solution. 


EXAMPLES. 


(1) Show that f “ody is the area bounded by a curve, the 
Y 


0 


axis of Y, and perpendiculars let fall from the ends of the 
bounding arc upon the axis of Y. 


(2) If the axes are inclined at the angle w, show that these 
formulas become 


bi Y 
A=sin of ydx = sin of ady. 
x Yo 


(3) Find the area bounded by the axis of X, the curve 
2 +4y=0, and the ordinate of the point corresponding to the 
abscissa 4. Ans. 54. 


(4) Find the area bounded by the axis of X, the curve 
y=, and the ordinates corresponding to the abscisse —2 
and 2. Ans. 8. 


(5) Find the area bounded by the axis of X, the axis of Y, 
the curve y=cos@, and the ordinate corresponding to the 
abscissa 37. Ans. 6. 


Cuap. X.] AREAS. 139 


126. In polar codrdinates we can regard the area between two 
radii vectores and the curve as the limit of the sum of sectors. 

The area in question is the sum 
of the smaller sectorial areas, any 
one of which differs from $7°A¢ by 
less than the difference between the 
two circular sectors $(7 + Ar)?Ad 
and 47°Ad; that is, by less than 





ae 
rArdd 4 ie , which is of the 


second order if Ad is the principal infinitesimal. 


limit [ ,?=%, ] 
Hence A= ; E rAd |, 
Ag = 0 oA > ? 
A=4] rd¢d 
% 


127. Let us find the area between the catenary, the axis of 
X, the axis of Y, and any ordinate. 


A= { ydx = Sf (et +e*) da, 
0 pe ae 
a4 ‘sree ie a 
culo (ea 6 a), 
but ; (ea — ea) aig by Art. 101. 


Hence A= as, 


and the area in question is the length of the arc multiplied by the 
distance of the lowest point of the curve from the origin. 


128. Let us find the area between the tractrix and the axis 
of X. 


We have dx =—-YVJG—%. (Art. 102.) 
y 


A = f yde = — | dyVa? — 7. 


140 INTEGRAL CALCULUS. [ART. 129. 


The area in question is 


0 2 
A=— (dyVa—yY=™, 


which is the area of the quadrant of a circle with a as radius. 


EXAMPLE. 


Give, by the aid of infinitesimals, a geometric proof of the 
result just obtained for the tractrix. 


129. In the last section we found the area between a curve 
and its asymptote, and obtained a finite result. Of course this 
means that, as our second bounding ordinate recedes from the 
origin, the area in question, instead of increasing indefinitely, 
approaches a finite limit, which is the area obtained. Whether 
the area between a curve and its asymptote is finite or infinite 
will depend upon the nature of the curve. 

Let us find the area between an hyperbola and its asymptote. 

The equation of the hyperbola referred to its asymptotes as 


axes is ea) w+ 
4 
Let w be the angle between the asymptotes ; then 


(oe) 2 9 ie 8) 
: a 2 Fer da 
A= sin 0 f ude me ae sin of —=~n 
0 4 ange 5 





Take the curve yue=4a?(2a—2), 


2a—2 
<P) aS eee 


or | afi Oe 
any value of # will give two values of y equal with opposite 
signs; therefore the axis of # is an axis of symmetry of the 
curve. 

When «= 2a, y=0; as a decreases, y increases; and when 
x=0,y=o. If is negative, or greater than 2a, y is imagi- 
nary. The shape of the curve is something like that in the 


Cuap. X.] AREAS. 141 


figure, the axis of Y being an asymptote. The area between the 
curve and the asymptote is then either 


A=2{ yd or A=2{xdy; 
0 


by the first formula, 


FAD Om 
A=4a PRa= = gy aa? . 
af \\ a x CU 1 


bf 





by the second, 


A= 16.08 fOr =4aq?r. 
oy +407 


EXAMPLES. 
(1) Find the area between the curve 9?(a? + a?) = a?2’ and its 
asymptote y= a. Ae A. eee 


(2) Find the area between 9°(2a—2)=2' and its asymptote 
= Us Ans. A =3-7a". 


2 > 
(3) Find the area bounded by the curve y?= ete) and 
its asymptote =a. Mune 


(— 


Maire 2a°(1 4 i): 


130. If the codrdinates of the points of a curve are ex- 
pressed in terms of an auxiliary variable, no new difficulty is 


presented. 
Take the case of the circle 2? + y? = a’, which may be written 





x= acosd : 
y=asing 
dy = acos ¢d¢. 


2 
The whole area A= at cos? ddd = ra’, 
0 


142 INTEGRAL CALCULUS. [ART. 131. 


EXAMPLES. 


(1) The whole area of an ellipse “= 7 8? ; is rab. 
y=bsind 


(2) The area of an arch of the cycloid is 37a’. 


(3) The area of an arch of the companion to the cycloid 
«= ad, y=a(1—cos6@) is 27a’. 


131. If we wish to find the area between two curves, or the 
area bounded by a closed curve, the altitude of our elementary 
rectangle is the difference between the two values of y, which 
correspond to a single value of x. If the area between two 
curves is required, we must find the abscissas of their points of 
intersection, and they will be our limits of integration ; if the 
whole area bounded by a closed curve is required, we must find 
the values of a belonging to the points of contact of tangents 
parallel to the axis of Y. 

Let us find the whole area of the curve 


aty’ + bat = a?b? 2’, 
or aty? = ba" (a?— 2”). 


The curve is symmetrical with reference to the axis of X, and 
passes through the origin. It consists of two loops whose areas 
must be found separately. Let us find where the tangents are 
parallel to the axis of Y. 


b 
y= =a Va — 2a’, 
a 


Be Sy Lg 
Gy PW Oe tenn 


da a? Va? petit 
T 


7 == — when tant = o, that is, when v=+ a. 





A=24 feve-# Pde +2 fa Va? = oF a. da =4ab. 


Cuap. X.] AREAS. 143 


Again ; find the whole area of (y —x)? =a? — 2. 
yea Rao Va? — 2, 
A ={w'- yl) des = ( 2VE =a? de 
To find the limits of integration, we must see where r= 5) 


dy _Ve@—a2 Fx 
Oa 


eel ° 
A=2| V@—2.de=7a?. 


= 


=oo when w=+ta. 


EXAMPLES. 


a? (a+ v7) 
ae 


Ans. 2a? ( = a 
4 


(2) Find the area between the curves y7—4az=0 and 
xe —day=0. | ne 16a? 


(1) Find the area of the loop of the curve 7° = 





(3) Find the whole area of the curve x? + yi=a?. Ans. 27a’. 
(4) Find the area of a loop of a’?y* = x*(a?— 2’). Ans. am 
(5) Find the whole area of the curve 
2y? (a? + 2) —4ay(a?— a’) +(e —2’yP=0. 
Ans. a’ (4 —_ ae) 
132. We have seen that in polar codrdinates 


$4 
pies bf dd. 
$o 


Let us try one or two examples. 
(a) To find the whole area of a circle. 


The polar equation is ee 


A= AG dob = 7a*. 


144 INTEGRAL CALCULUS. (ART. 132. 


(®) To find the area of the cardioide 7 = 2a(1 — cos¢). 


ic £ (4c —cos¢)*d¢= 2at (1 2cos hd + cos’dd) dd, 
A=6a'r. 
(c) To find the area Butreon an arch of the epicycloid and the 
circumference of the fixed circle. 
2 =(a+b)cos6 —b cos# +" =p 
= a, 


We can get the area bounded by two radii vectores and the 
arch in question, and subtract the area of the corresponding 
sector of the fixed circle. 

Changing to polar coérdinates, 





y =(a+5b)sin 0 —bsin 


C= 9 COS wD, 
y=rsin ¢d. 
We want bf rag. 
tan @ == 3 
dy— ydx 
sec? ddd =~ 
bg = ot 
but, since Y=Prcosd, seco= af : 
z 
hence ree Sat tesae 
xe fn) 
and rdd = audy— ydz ; 


da = (a +)(— sin 6 +sin 2? +) a, 





dy =(a +2) (cos — cos *6) a 
wdy — ydx=(a+b)(a+ 20y(1 + cos 0)a0 = dd. 


Our limits of integration are obviously 0 and 20 
a 


Cuar. X.] AREAS. | 145 


2677 
Hence A=4(a+bd)(a+2 afr _ cos 8) dé, 
yo G a by (a + 2b), 
a 


is the area of the sector of the epicycloid. Subtract the area of 
the circular sector 7ab, and we get 


2 
Aa PBat 20) 


a 
as the area in question. 


(d) To find the area of a loop of the curve 7” = @’cos2 ¢. 

For any value of ¢ the values of 7 are equal with opposite 
signs. Hence the origin is a centre. 

When ¢=0, r=+4; as @ increases, 7 decreases in length 


: T T : : : > 
till 6= 7 when 7 = 0; as soon as ¢> re ris imaginary. If ¢ 


decreases from 0, 7 decreases in length until ¢= a wy Heri a == 0: 
and when ¢< 7 y is imaginary. To get the area of a loop, 
then, we must integrate from d= — 7 to ¢= 7 
T 
2(t a 
Jee {tras = bat cos2 d.dd = ae 
-- + 4 
EXAMPLES. 
| (1) Find the area of a sector of the parabola r= Sb AB Al 
1+ cos¢ 
(2) Find the area of a loop of the curve 7*cos¢ = a’sin3 ¢. 
2 2 
NS OM een 
4 2 
(3) Find the whole area of the curve 7 = a(cos2¢+ sin2 ¢). 
Ans. 7a’. 


(4) Find the area of a loop of the curve rcos¢ = a cos2 ¢. 


Ans. (2 — | a’. 


(5) Find the area between r =a(secd+tan¢@) and its asymp- 
tote rcos¢= 2a. yom e if 2) at, 


146 INTEGRAL CALCULUS. (Arr. 133. 


133. When the equation of a curve is given in rectangular 
codrdinates, we can often simplify the problem of finding its area 
by transforming to polar codrdinates. 

For example, let us find the area of 


(2+ y)?=40Ce?+407y’. 
Transform to polar coérdinates. 
m= 47°(a? cos’? d + sin’ ¢), 
7” = 4 (a? cos’ + 0b’ sin’), 


27 
He 2 { (arcos*¢ 4b? sin? p) deb = 27 (a? +02). 
0 


EXAMPLES. 
(1) Find the area of a loop of the curve (a+ y°)’=4 aay’. 


‘Ans. = a 
8 


2 2 2 
(2) Find the whole area of the curve —+ 7 = re +5) ‘ 


os 
A ° ae as b? ° 
i 2ab (4 Sage 
(3) Find the area of a loop of the curve y° — 3 avy +2°= 0. 


2 
Ans. 3a" 
2 


134. The area between a curve and its evolute can easily be 
found from the intrinsic equation of the curve. 

It is easily seen that the area 
bounded by the radii of curvature 
at two points infinitely near, by 
the curve and by the evolute, dif- 
fers from $p’dr by an infinitesimal 
of higher order. The area bounded 
by two given radii vectores, the 
curve and the evolute, is then 


un a, 
Ment a 2dr. 





Cuap. X.] AREAS. 147 


as T; ds 2 
Hence A= f a dr. 


For example, the area between a cycloid and its evolute is 


: . 2 
A =3f Cae or dh 
rant dlr 
0 
; oe 
SEs a? {cos 7dr. 


Let m=0 and 1=°-; 


2 b 

% 

A= 8a? { Fcos*rde = 27. 
0 


EXAMPLES. 
(1) Fin the area between a circle and its evolute. 
(2) Find the area between the circle and its involute. 


Holditch’s Theorem. 


135. Ifa line of fixed length move with its ends on any curve 
which is always concave toward it, the area between the curve 
and the locus of a given point 
of the moving line is equal 
to the area of an ellipse, 
of which the segments into 
which the line is divided by 
the given point are the semi- 
Axes. 

Let the figure represent 
the given curve, the locus 
of P, and the envelope of the 
moving line. 

Let dP=a and PB=), 
and let CB =p, C being the 
point of contact of the moving line with its envelope. Let 
AB=a+b=c. 





148 INTEGRAL CALCULUS. [ART. 135. 


The area between the first curve and the second is the area 
between the first curve and the envelope, minus the area between 
the second curve and the envelope. 

Let 6 be the angle which 
the moving line makes at 
any instant with some fixed 
direction. - Let the figure 
represent two near positions 
of the moving line; A@, the 
angle between these posi- 
tions, being the principal in- 
finitesimal. 

PB=p, P'B'=p+Ap. 


The area PBB'P'P differs 
from 4p°d@ by an infinitesi- 
mal of higher order than the first. 

4°d6 is the area of PBMP, and differs from PP'NB by less 
than the rectangle on PM and PQ, which is of higher order than 
the first, by I. Art. 153. But PP'NB differs from PP'B'B by 
less than the rectangle on BN and NB’, which is of higher order 
than the first, since VB’, which is less than PP'+ Ap, is infini- 
tesimal and A@ is infinitesimal. 

The area between the first curve and the envelope 1s then 


om 


+(% p' dé; or, since we can take PP'A'A just as well for our 
0 
elementary area, fe (¢ — p)? dé. 

0 


Tv 27 
Hence 4 (pag to ANG —p)°dé; 
T 





whence 2 ef pas bs a 
27 
or f ode es ee! (1) 
0 


The area between the second curve and the envelope is 


27 
£(o- b)2d6. 
Q 


Cuap. X.] AREAS. 149 


The area between the first curve and the second is then 


2G a2 
A=4[p'd6—3 )(p — 0)? 


oT 
= { pd — bx 
) 


= rbe — br by (1), 
=7b(a+b)—0'7, 
A=7ab, (2) 


which is the area of an ellipse of which a and 0 are semi-axes. 
Q. E. D. 


EXAMPLES. 


(1) If a line of fixed length move with its extremities on two 
lines at right angles with each other, the area of the locus of a 
given point of the line is that of an ellipse on the segments of 
the line as semi-axes. 


(2) The result of (1) holds even when the fixed lines are not 
perpendicular. 


Areas by Double Integration. 


136. If we take x and y as the codrdinates of any point P 
within our area, x and y will be independent variables, and 
we can find the area bounded by two 
given curves, y= fe and y = F2, 
by a double integration. Suppose 
the area in question divided into 
slices by lines drawn parallel to the 
axis of Y, and these slices subdi- 
vided into parallelograms by lines 
drawn parallel to the axis of X. 
The area of any one of the small 
parallelograms is AyAw. If we 
keep x constant, and take the sum 
of these rectangles from y= fx to y = Fx, we shall get a result 
differing from the area of the corresponding slice by less than 





150 INTEGRAL CALCULUS. [ART. 137. 


2 AvAy, which is infinitesimal of the second order if Ax and Ay 
are of the first order. 
Fx Fu 
Hence f. Az.dy = Ax | dy 
fx fx 

is the area of the slice in question. If now we take the limit of 
the sum of all these slices, choosing our initial and final values 
of x, so that we shall include the whole area, we shall get the 
area required. 


Hence S A={" ( if iy) dx. 


In writing a double integral, the parentheses are usually omit- 
ted for the sake of conciseness, and this formula is given as 


xz, (ke 
ae ah J dude, 


the order in which the integrations are to be performed being the 
same as if the parentheses were actually written. 

If we begin by keeping y constant, and integrating with respect 
to x, we shall get the area of a slice formed by lines parallel to 
the axis of X, and we shall have to take the limit of the sum of 
these slices varying y in such a way as to include the whole area 
desired. In that case we should use the formula 

CP pape 
JA a dady. 
tes iat 


137. For example, let us find the area bounded by the para- 
bolas y? = 4a and 2 = 4ay. 
The parabolas intersect at the origin and at the point (4a, 4a). 


4a (7° 4ax 4a (7~4ay 

A= rf er or A=f{ aid 
J hax 2 

Lee V4ax —= ; 

4a 


4a (V4axr 4a 
anh aes ={ Aas dx = V6 ge. 
0 4a 3 


The second Zslsninla gives the same result. 








Cuap. X.] AREAS. 151 


EXAMPLES. 


(1) Find the area of a rectangle by double integration ; of a 
parallelogram ; of a triangle. 


(2) Find the area between the parabola 7’? = ax and the circle 


y= 200 — a. Ans. 2 wa 2 ay. 
4 3 

(3) Find the whole area of the curve (y — mx —c)* =a? — 2’. 

Ans. 7a’. 


138. If we use polar coordinates we can still find our areas 
by double integration. 

Y Let r=f¢ and r= F'¢ 
be two curves. Divide the 
area between them into 
slices by drawing radii 
vectores; then subdivide 
these slices by drawing 
ares of circles, with the 
origin as centre. 

Let P, with codrdinates 
yr and @¢, be any point 
within the space whose 
area is sought. The curvilinear rectangle at P has the base rAd 
and the altitude Ar; its area differs from rA¢Ar by an infinitesi- 
mal of higher order than rA¢Ar. 





Ke = Pe 
The area of any slice as aba’b' is f rAddr, ¢ and Ad being 
So 


7) 
constant, that is Ag { rar. The whole area, the limit of the 


So ¢, pF 
sum of such slices is d= f rdrdd. (1) 
% “SP 


Or we may first sum our rectangles, keeping 7 unchanged, 
and we get as the area of efe'f' 


F- ‘6 
rAr fas. and A =f J rasa. (2) 


152 INTEGRAL CALCULUS. TART. 138. 


It must be kept in mind that 7 in (1) and (2) is the radius 
vector of any point within the area sought, and not of a point 
on the boundary. 


For example, the area between two concentric circles, r= a 
and r= J, is 


a 27 27 a 
A ={ fragar ={ [rarag = 7(a?— 0’). 
b 0 0 Jt 


Again, let us find the area between two 
tangent circles and a diameter through the 
point of contact. 
Let a and 0 be the two radii, 
r= 2acos¢d eo) 
and r=2bcos¢d (2) 
are the equations of the two circles. 


~ 


={ * (mands = en (4 -) (bos odd = = 5 (@ — 6°). 


bcos¢ 


If we wish to reverse the order of our integrations we must 
break our area into two parts by an arc described from the origin 
as a centre, and with 2b as a radius; then we have 


Z eit ‘fer Hie em 
Radi pe 10s"! —cos- dr + oe 
={r cos”'57 ! et 0s 5 qt 


— 5h Be): 


EXAMPLE. 


Find the area between the axis of X and two coils of the 
spiral += ad. 


Cuap. XI] AREAS OF SURFACES. 1538 


CHAPTER XI. 
AREAS OF SURFACES, 


Surfaces of Revolution. 


139. If a plane curve y= fz revolves about the axis of X, the 
area of the surface generated is the limit of the sum of the areas 
generated by the chords of the infinitesimal 
arcs into which the whole are may be broken 
up. Each of these chords will generate the 
surface of the frustum of a cone of revolution 
if it revolves completely around the axis; 
and the area of the surface of a frustum 
of a cone of revolution is, by elementary ~ 
Geometry, one-half the sum of the circum- 
ferences of the bases multiplied by the slant height. The frustum 
generated by the chord in the figure will have an area differing 
by infinitesimals of higher order from z(y+y-+Ay)As or from 
2ryds. The area generated by any given arc is then 





aT 
S=27] yds. [1] 
Yo 


If the arc revolves through an angle 6 instead of making a 
complete revolution, the surface generated is 


Y 
S= af yds. [2] 


It must be noted that [1] and [2] will give a positive value 
for S if the generating curve lies wholly above the axis of X at 
the start, and a negative value for S if it lies wholly below the 
axis of X at the start. If the curve happens to cross the axis 
of X between the points whose ordinates are 'y) and y,, [1] and 
[2] give not the area of the surface generated by the curve in 
question, but the difference between the areas generated by the 


154 INTEGRAL CALCULUS. (Arr. 140. 


portion originally above the axis, and the portion originally 
below the axis. 
EXAMPLE. 


Show that if the arc revolves about the axis of Y, S= 2x f ads. 
% 


140. To find the area of a cylinder of revolution. 
Take the axis of the cylinder as the axis of X. Let @ be the 
altitude and 0 the radius of the base of the 
cylinder. The equation of the revolving 
line is 

y=; 

dy = 0, 

ds = Vda? + dy’ = da; 


S=2r "yale = 2ab, 





or the product of the altitude by the circumference of the base. 
Again, let us find the surface of a zone. 
The equation of the generating circle is 





Aya +7 po "1 ee 
ada 
ds = ; 
y 


Se anf ada = 2am (2 — %%). 
Lo ay ; 
If m%=—a and 1,=a, S=4a’r. 
Hence the surface of a zone is the altitude of the zone multi- 
plied by the circumference of a great circle, and the surface of 
a sphere is equal to the areas of four great circles. 
Again, take the surface generated by the revolution of a 


cycloid about its base. 


, 


x= abd —asing 
y=a—acosé 


ds = ad6~/2 (1 — cos8), by Art. 105; 
20 5 

S=2nf a*./2.(1 — cos 6)? dé = 54A7a?, 
0 


Cap. XI.] AREAS OF SURFACES. 155 


EXAMPLES. 


(1) The area of the surface generated by the revolution of 
the ellipse te ey 





about the axis of X is 27ab ees et sin7 )) 





about the axis of Y is 27a? fa + see “loe2 oe = 
—e 


a? — b? 


9 


4 


where e?= 


(2) Find the area of the surface generated by the revolution 
of the catenary about the axis of X; about the axis of Y. 


(3) The whole surface generated by the revolution of the 
tractrix about its asymptote is 47a’. 


(4) The area generated by the revolution of a cycloid about 
its vertical axis is 87a?(z— 4). 


(5) The area generated by the revolution of a cycloid about 
the tangent at its vertex is 32-7a’. 


(6) The area generated by the revolution of the curve 
a + y? =a? about its axis is 1270. 


141. If we know the area generated by the revolution of a 
curve about any axis, we can get the area generated by the 
revolution about any parallel axis by an easy transformation of 
codrdinates. 

Given the surface generated by the arc from s) to s, about 


8 é OX, to find the area generated by 
: a the same arc when it revolves 
Paes x' about O'X'. 


Let S be the surface about OX, 
and S' about O'X', 


| i We have 


8, 3; 
S=2r{fyds, S'’=2r { y'ds’. 
50 


So 


156 INTEGRAL CALCULUS. (ArT. 141. 


By Anal. Geom.. 22. 
y=yty. 
Hence dz=dz', dy=dy', ds=ds', 


and S= 2= | (y+y')ds= 2 Yo( 1 — So) + 2z yds, 
So So 


= 2 rU(S 4 8) _ S". 
Therefore S'= S$ — 27y(s, —%). [1] 


$,— S is the length of the revolving curve; 2zy is the cir- 
cumference of a circle of which y is the radius. Hence the new 
area is equal to the old area minus the area of a cylinder whose 
length is the length of the given arc and whose base is a circle 
of which the distance between the two lines is radius. 

In using this principle careful attention must be paid to the 
sign of yw. and it must be noted that the original formula 


os : 
S=22 f yds will always give a negative value for the area of 
So 


the surface generated. if the revolving arc starts from below the 
axis ; and hence, that the surface generated 
by the revolution of any curve about an 
axis of symmetry will come out zero. 

As an example of the use of the princi- 
ple, let us find the surface of a ring. 

Let a be the distance of the centre of OD x 
the circle from the axis, and 4 the radius of 
the circle. Since the area generated by the 
revolution of the circle about a diameter is zero, the required 
area is 


EXAMPLE. 
Find the area of the ring generated by the revolution of a 
cycioid about any axis parallel to its base. 


7 Pray dS: sabe (= + sor) 





vy. eas 


—. 


Cuap. XI.] AREAS OF SURFACES. 157 


142. If we use polar codrdinates, 


3; 
S= 2x f yds 
. 


0 


3; 
becomes S=2r if r sin d.ds. 
30 
where ds = Vd7* + 7°d¢’. 


For example ; let us find the area of the surface generated by 
the revolution of the upper half of a cardioide about the hori- 
zontal axis. 

7 = 2a(1—cos¢); 
dr = 2asind.dd, 
ds’ = 8a?(1—cos¢)d¢’, 
= 2a {4 J2eA- cos $) ‘sin ¢.d¢. 
0 


S = 128 7a’. 


EXAMPLES. 


(1) Find the surface of a sphere from the polar equation. 


(2) Find the surface of a paraboloid of revolution from the 
polar equation of the parabola 


ae m 
1—cos¢@ 


Cylindrical Surfaces. 


143. If a cylindrical surface is generated by a line which is 
always parallel to the axis of Z, the area of the portion bounded 
by two positions of the generating line, the plane of XY, and 
any curve whose projection on the plane of XZ is given, is 
easily found. 

Let ABCD be the cylindrical area required. 


158 INTEGRAL CALCULUS. [ART. 143. 


Let y = fu (1) 


be the equation of AB, the line of intersection of the surface 
with the plane XY; and let 


Bi fie (2) 


be the equation of C,D,, the projection of CD on the plane 
of XZ. 

If x,y,z are the codrdinates of 
any point P of CD, the required 
area is evidently the limit of 
the sum of rectangles, of which 
PP'P"P" is any. one. -‘The area 
of PP'P"P" differs by an in- 
finitesimal of higher order than 
ds from zds, and therefore the 


cf 
required area S = if zds. 
x 





x,z are the coordinates of P,, and 
satisfy (2), and ds = Vda? + dy? 
where 2,y are the codrdinates of 
P' and satisfy (1). 





We have, then, S= f ” Vd? + dy’. [3] 


0 


For example, let AB be the quadrant of a circle, and let the 
projection of the required area on the plane of XZ be the quad- 
rant of an equal circle, so that the surface required is one-eighth 
of the surface of a groin. © 


Here e+y= a’, (4) 
and e+e2=a*; (5) 


da 
ds = Vda? + dy? = de = “fF 
y Va? — oe 


and 2= Va?— 2’. 


Cap. XI.] AREAS OF SURFACES. 159 


Therefore S= fl bed bey ae eee ‘do a 
0 Va? — x 0 


Again, let us find the area of the curved surface of the 
portion of a cylinder of revolution included within a spherical 
surface, whose centre lies on the surface of the cylinder, and 
whose radius is equal to the diameter of the cylinder. 

If the centre of the sphere is taken as the origin, and a 
diametral plane of the cylinder as the plane of XZ, the surface 
required is four times that indicated in the figure. 

The equiition. of the cylinder is 


v—ax+y’?=0, (6) 
and of the sphere 
ety+te—a=0. (7) 
Subtract (6) from (7), and we get 
2+ax—a’= 0 (8) 





as the equation of a cylindrical surface 
perpendicular to the plane XZ, and 
passing through all the points of intersection of (6) and (7). 
(8) is, then, the equation of the projection on the plane of XZ 
of the line of intersection of the given spherical surface and 
the given cylindrical surface. 


From (6), ds = Vda?+dy= SL ea ats 
} 2y 2Vaxe — a? 
From (8), z= Va?—az. 
Hence he Ber CES ft yar “Va — 2. de 
9 2Var—a gee Vava— a 
“ae = a’; 
VE 
and the whole area required, 


4S= 4a’, 


160 INTEGRAL CALCULUS. (ArT. 144. 


EXAMPLES. 


(1) Find the area cut from the cylindrical surface whose 
base in the plane XY is a quadrant of the curve wv + yi = a by 
the plane «=z. Ans. 2a’. 


(2) Find the area of that portion of a cylindvical surface 
mpese base in the plane of XY is a quadrant o: the ellipse 


2 


eo ie at+5 “—=1, and whose projection on the plane of XZ is bounded 
Vag ten b(@ + ab + 0") 
3(u +b) 


(8) Let the base of the cylindrical surface be a tractrix, 
whose vertex lies at a distance a to the left of the origin, and 
whose asymptote is the axis of Y, while its projection on the 
plane of XZ is bounded by the parabola 2 = —2 mz. 


Ans. S= 3aV2ma. 


ie ne curve a?y’ = b?a?(a? — 2”). 


(4). Let the base of the cylindrical surface be the upper half 
of a cycloid, having its vertex at the origin and its base parallel 
to the axis of Y, and at a distance 2a from the origin, while 
its projection on the plane of XZ is bounded by the parabola 


y? = 2 mx. Ans. S=4avam. 


Any Surface. 


144. Let a, y, z be the codrdinates of any point P of the sur- 
face, and a+Aa, y+Ay, z+ Az the coodrdinates of a second 
point Q infinitely near the first. Draw planes through P and Q 
parallel to the planes of XY and YZ. These planes will inter- 
cept a curved quadrilateral PQ on the surface ; its projection pq, 
a rectangle, on the plane of XZ; and a parallelogram p'q' not 
shown in the figure, on the tangent plane at P, of which pq is 
the projection. PQ will differ from p'g' by an infinitesimal of 
higher order, and therefore our required surface will be the limit 
of the sum of the parallelograms of which p’‘q' is any one. 


Cuap. XI.] AREAS OF SURFACES. 161 


If 8 is the angle the tangent plane at P makes with XZ, 
p'g' cosB=pq or p'q'=pqgsecB = AwAzsecB, and o, our sur- 
face required, is equal to 
the double integral 


oe fs if sec Bdadz 


taken between limits so 
chosen as to embrace the 
whole surface. 

The limit of the sum 
of the parallelograms, of 
which p’g’ is a type, will 
be the required surface 
if the limit of the sum of 
the rectangles, of which 
pq is a type, is the pro- p 
jection of the surface in 
question on the plane of XZ; so that the values of 2 and » 





between which we integrate in o= f { sec Bdadz are precisely 

those we should use if we were finding the area of the projection 

of o by the double integration f f dadz. (v. Art. 136.) 
The equation of the tangent plane at P is 

(2% — a) Do, F+(Y — Y)DyF + (% — %) Da f=9, by I. Art. 217, 

(25 Yos Z) standing for the codrdinates of the point of contact, 

and f(x,y,%)=0 being the equation of the surface. 


The direction cosines of the perpendicular from the origin upon 
the plane are 
P Dy f 


Oe nae! Seo Sr sae 
V (Da, fY + (Dy SP + (Dz SP 
cos B = po eee Oe 
eet f- (Dy, f° ar (Dz, f) 
aoe Dz, f 
V (Daf) + (Dy SY + (Daf)? 


by Anal. Geom. of Three Dimensions. 








COS Y 





162 INTEGRAL CALCULUS. _ ArT. 144. 


Hence, dropping the accents, 


c= { (G22 OC) Oia [1] 
TENE 


By considering the projections upon the other codrdinate planes 
we shall find 





i f f= (Df)? + (DFP + (DAY ayag 2] 
a : WBA § 

net {paueaiaae (Df)? Ds (DFP +OSY oay, 3] 
D.f 


In each of the formulas the derivatives are partial derivatives. 
Let us find the area of the portion of the surface of the sphere 


e+ypPt ea 


intercepted by the three codrdinate planes. 


DOM byt 
D, f= 2Y, 
YBN, Pw 4 


V(D,f)? + (D,f)?-+ (Df)? = 2a. 


Sate 22 


o at ‘fs dydz ; (1) 
Sa2— 22 
or o={ le dzdx ; (2) 
or o =f 2 dxdy. (3) 
0 Joz 


For, in the second one, which agrees best with the figure, we 
must take our limits so that the limit of the sum of the projec- 
tions may be the quadrant in which the sphere is cut by the 


Crap. XI.] AREAS OF SURFACES. 163 


plane XZ; and the equation of this section is obtained by letting 
y = 0 in the equation of the sphere, and is 





whence g= Va? — 2. 
9 


If we take as our limits in the integral {) “dz zero and Ve— & 


we shall get, the area whose projection is a strip running from 
the axis of Z to the curve; then, taking { ( f “ dz) da from 0 to 
y 


a, we shall get the area whose projection is the sum of all these 
strips, and that is our required surface. 


y= VO 2 2, 


ieee OP dade 
o= a 
oa aS RE. — a? 


(= = tear 
Va? — a? — 2? Va? — a 


if we regard x as constant ; 








a0 ra 
o=af -dvi=—, 
0 2 2 


the required area. Formulas (1) and (3) give the same result. 


145. Suppose two cylinders of revolution drawn tangent to 
each other, and perpendicular to the plane of a great circle of a 
sphere, each having the radius of the 
great circle as a diameter ; required the 
surface of the sphere not included by 
the cylinders. 

The surface required is eight times 
the surface of which the shaded portion 
of the figure is the projection. 

If we take the plane of the great 
circle as the plane of XY, 





164 INTEGRAL CALCULUS. (ArT. 145. 


wo —ar+ y= 0 (1) 
is the equation of the cylinder, and 


e+y+e= 0? (2) 
of the sphere. 


We have = eae 


Dif 

From (2) Do Fises ase 
D,f = 2y; 

D, f = 22; 


(Df)? + (Df)? + . fyr=4a’. 


Hence c= =i f; dydx =a f a 
Vei—x#—¥ 


Our limits of integration for y are Vaw — 2 and Va? — 2°; for 
« are 0 and a. 





Max dydx 
oes af Sofas 
Jax —x? 
J a2 x2 Re 
JS a2— x2 dy Waa y ape ; -14] 
Jee a SR a+2 
Jax — x? Vax —x2 


.dx we must integrate by parts. 





To find if ated 
0 








: x 
Let u= sin-*| . 
a+n2 
and dv = dz; 
V= 2, 


Qt = A eae. 
2(a+a) Va 


fain a dy = sin] _Na Va 2% 
ate ate 2/7a+2 











Cuap. XI.] AREAS OF SURFACES. 165 








Let Wa=n/e; 2wdw= dx 
s/e.da _ wedw fi 
and a+2 mae a+ w it es ane dw. 





ede of ss 
f= w — /a tan va ; 


: /2 (ltr a7 ar 
SUP Sitke eee 1 CAT 1 — Oh oe of 
2 4 io 4 2 i 


a7 Car 
paa(S_S +a) = q?. 





8c = 8a’ is the whole surface in question. 


146. Let us find the area of the curved surface of a right 
cone whose base is the curve #3 + y= a3, and whose altitude 
Ge 

If we take the vertex of the cone as the origin of codrdinates, ~ 
and its axis as the axis of Z, the equation of its curved surface 


is 
ot yt = (=) (1) 


and the projection of the surface on the plane of X Y is bounded 


by the curve 
xe + yi = al. (2) 


From (1) we get 


V (DS) + Gof mae 0 ae a+ (wt + yt) 


Ayes ys 





$s 
where «#,y are the codrdinates of any point within the projec- 
tion of the base of the cone. 
Since the four faces of the cone are equal, the peduireg 
surface 
(ata)! 


c= { a tyt Var at yt + 2 (ai + y8)?. dy de. (3) 


166 INTEGRAL CALCULUS. fArr. 146. 


Let us substitute v’=a and w*=y, whence dxa=3v'?dvu 
and dy=3w’dw, and we have 


at (at_v2)t 


c= Tf few Varv2w? + (v2 +w'). dwdv ; 
0/0 


or, since in a definite integral it makes no difference what letters 
we use for the variables, 


at. (aé_x?) 
re A Voraty? + (a? +7). dy da. (4) 
The « and y in (4), however, must not be confounded with the 
xand y in (3). 

The integral in (4) is precisely that which we should have to 
find if we sought the area of a surface of such a nature that its 
projection on the plane of X Y was a quadrant of the circle 
x+y? =a, and the secant of the angle made by the tangent 
plane at any point (#,y,z) of the surface with the plane of X Y 
was xy Varay? + (a? + yy. 

In the latter problem there is nothing to prevent our re- 


Ons 





placing # and y in ay Vara?y? + c (a? +4)? by their values in 
terms of r and q, the polar codrdinates of any point of the 
projection 2? +7?= ai, and dividing this projection into polar 
elements instead of rectangular elements, and then integrating 
between the limits which we should use if we were finding the 


area of the projection by the formula A =f ragar. 


We have, then, 


ot ig | 
c= Bt sind cos Va?r* sin? d cos’? + 27*. rdrdd 
Ob 0 ry Mi b] 


or 


BB (fine A/a sin? comb oe nae 
c= Tf hi sin d cOS@ Va’ sin’ d cos*¢ +c’. drdd, 


r| 4a 





Ta ba f sind cos ¢ Va? sin? cos’ +¢°. dd. 
0 


Cuap. XI.] AREAS OF SURFACES. 167 


Substitute w—=sin?d, and 


al Penny. AnOite hits be ee 
o= Saf Vatu (1—u) +e. du, 
e/) 
c= af? ac + (c? + 4a") tan“? a 
aye 


EXAMPLES. 


(1) Find the area included by the cylinders described in 
Art. 145 by direct integration. 


(2) A square hole is cut through a sphere, the axis of the 
hole coinciding with a diameter of the sphere; find the area of 
the surface removed. 


(3) A cylinder is constructed on a single loop of the curve 
r=acosnd, having its generating lines perpendicular to the 
plane of this curve; determine the area of the portion of the 
surface of the sphere w+ y’+2*=a? which the cylinder inter- 
ae: Ans. iG 

n \2 

(4) Find the area of the portion of the surface of the cone 

described in Art. 146 included by the cylinder 2 + 7° = 0’. 


Ans. ac VaeI+3e tan (“248 = —@ tant 
i G 





, Cc 
(5) Find the area of the portion of the surface of the sphere 
ert+yt2z=2ay cut out by one nappe of the cone 
Ax’? + B2=y’. Ane Ara? 
Vy (A) (PB) 
(6) Find the area of the portion of the surface of the sphere 
e+ y?+27=2ay lying within the paraboloid y = Ax’ + Bz’. 
Ans. 210 
: AB 
(7) The centre of a regular hexagon moves along a diameter 
of a given circle (radius = a), the plane of the hexagon being 
perpendicular to this diameter, and its magnitude varying in 
such a manner that one of its diagonals always coincides with 
a chord of the circle ; find the surface generated. 
Ans. 0 (27+3,/3). 








168 INTEGRAL CALCULUS. TARY. 147, 


CHA PACE Tt os Ui. 
VOLUMES. 
Single Integration. 


147. If sections of a solid are made by parallel planes, and a 
set of cylinders drawn, each having for its base one of the sec- 
tions, and for its altitude the distance between two adjacent 
cutting planes, the limit of the sum of the volumes of these 
cylinders, as the distance between the sections is indefinitely 
decreased, is the volume of the solid. 

We shall take as established by Geometry the fact that the 
volume of a cylinder or prism is the product of the area of its 
base by its altitude. 

It follows from what has just been said, that if, in a given 
solid, all of a set of parallel sections are equal, the volume of 
the solid is its base by its altitude, no matter how irregular its 


form. 
Let us find the volume of a pyramid having 0 


for the area of its base, and « for its altitude. 
Divide the pyramid by planes parallel to the 
base, and let z be the area of a section at the dis- 
tance « from the vertex. 
We know from Geometry that 


Hence Z= oe : 
a 
Let the distance between two adjacent sections be dx; then 
the volume of the cylinder on z is 
2 wax, 
we 
and V, the required volume of the pyramid, is. 


h 


9 
ap 


Z 
bb’ oF 








Crap. XII. ] VOLUMES. 169 


Precisely the same reasoning applies to any cone, which will 
therefore ‘have for its volume one-third the product of its base 
by its altitude. 

EXAMPLE. 


Find the volume of the frustum of a pyramid or of a cone. 


148. If a line move keeping always parallel to a given plane, 
and touching a plane curve and a straight line parallel to the 
plane of the curve, the surface generated is called a conoid. 
Let us find the volume of a conoid when the director line and 
curve are perpendicular to the given plane. 

Divide the conoid into laminae by 
planes parallel to the fixed plane. 

Let Ay be the distance between 
two adjacent sectiens, and let a be 
the length of the line in which any 
section cuts the base of the conoid ; 
let a be the altitude and b the area 
of the base of the figure. Any one of our elementary cylinders 
will have for its volume }axAy, since the area of its triangular 





base is ax, and we have V=saf xdy, the limits of integra- 


tion being so taken as to embrace the whole solid. de xdy he- 


tween the limits in question is the area of the base of the co- 


noid; hence its volume, 
ie = 3 ab. 


EXAMPLES. 


(1) Find the volume of a conoid when the director line and 
curve are not perpendicular to the given plane. 


(2) A woodman fells a tree 2 feet in diameter, cutting half- 
way through from each side. ‘The lower face of each cut is 
horizontal, and the upper face makes an angle of 45° with the 
horizontal. How much wood does he cut out? 


170 INTEGRAL CALCULUS. [Arr. 149. 
149. To find the volume of an ellipsoid. 


=i 


faa rie gs 
ae bh? lod 


Take the cutting planes parallel to the plane of XY. A sec- 
tion at the distance z from the origin will have 





ge et rae IE big 
AER ‘be ¢ 
. Lk BOREAICE 2 
for its equation, and — Vc? — 2 and — ° Jee c’ — 2 for its semi-axes ; 
: : rab 
hence its area will be ——(c’?—2’). 
a 


Any of the elementary cylinders will have for its volume 


ow — z°)Az, and we shall have for the whole solid 








Pee aoe fie 2) da. 


V = §rrabe. 
If a, b, and ¢ are equal, the ellipsoid is a sphere, and 


Vieee aa: 


EXAMPLES. 


(1) Find the volume included between an hyperboloid of one 
sheet 


2 2 
Bras) apie eae 
a te A Ou 1 4 
Cr as 

and its asymptotic cone 
| ae gM ee 

whe 79 Tg Gack 
ODN 


Ans. It is equal to a cylinder of the same altitude as the 
solid in question, and having for a base the section made by the 
plane of XY. 


(2) Find the whole volume of the solid bounded by the surface 





y 
{+4 ce of = =] ° Ans. 8 zi 


Cuap. XII. ] VOLUMES. 1th 


(3) Find the volume cut from the surface 


by a plane parallel to the plane of (YZ) at a distance a from it. 
Ans. ra?,/ (bc). 


(4) The centre of a regular hexagon moves along a diameter 
of a given circle (radius = a), the plane of the hexagon being 
perpendicular to this diameter, and its magnitude varying in 
such a manner that one of its diagonals always coincides with 
a chord of the circle; find the volume generated. 

Ans. 2./3.a°. 

(5) A circle (radius = a) moves with its centre on the cir- 
cumference of an equal circle, and keeps parallel to a given 
plane which is perpendicular to the plane of the given circle ; 


find the volume of the solid it will generate. 2 a8 
Ans. Sar +8). 


Solids of Revolution. ‘Single Integration. 


150. Ifa solid is generated by the revolution of a plane curve 
y = fx about the axis of x, sections made by planes perpendicu- 
lar to the axis are circles. The area of any such circle is wy’, 
the volume of the elementary cylinder is 7y*Aw, and 


» 


Ae pda 
% 
is the volume of the solid generated. aa 
For example ; let us find the volume of the solid generated by 
the revolution of one branch of the tractrix about the axis of X. 
Here we must integrate from «= 0 to v=o. 


Ver ( ye. 
0 


(a? —y?)} 
We have dx = — eye dy (Art. 102 [2].) 
in the case of the tractrix ; 


172 INTEGRAL CALCULUS. (Arr. 151. 


hence V=— fue — y’)idy. 
= 
When 2=0, y= a, and whenia =|, y= 0. 
0 i) 
Therefore V=— f y(a — xy’) idy = a 
EXAMPLES. 


(1) If the plane curve revolves about the axis of Y, 


V=qr f Ny dy. 
A) 


(2) The volume of a sphere is 4 za’. 


(3) The volume of the solid formed by the revolution of a 
cycloid about its base is 57° a’. 


(4) The curve ¥°(2a — x) = 2 revolves about its asymptote ; 
show that the volume generated is 27°’. 


(5) The curve 2+ yi=a? revolves about the axis of X; 
show that the volume generated is 2% 7a’. 


Solids of Revolution. Double Integration. 


151. If we suppose the area of the revolying curve broken up 
into infinitesimal rectangles as in Art. 125, the element AxvAy 
at any point P, whose codrdinates are w and y, will generate 
a ring the volume of which will differ from 2mryAwAy by an 
amount which will be an infinitesimal of higher order than the 
second if we regard Ax and Ay as of the first order. For 
the ring in question is obviously greater than a prism having 
the same cross-section AwAy, and having an altitude equal to the 
inner circumference 2zy of the ring, and is less than a prism 
having AwAy for its base and 27(y + Ay), the outer circumfer- 
ence of the ring, for its altitude ; but these two prisms differ by 
27Aa(Ay)’, which is of the third order. 


Cuap. XII.] VOLUMES. 173 


: | 
Ax | 2zydy, where the upper limit of integration is the ordi- 


nate of the point of the curve immediately above P, and must be 
expressed in terms of x by the aid of the equation of the revolv- 
ing curve, will give us the elementary cylinder used in Art. 137. 

The whole volume required will be the limit of the sum of 
these cylinders ; that is, | 


nas 2m ff ydyae. [1] 
x0 0 


If the figure revolved is bounded by two curves, the required 
volume can be found by the formula just obtained, if the limits 
of integration are suitably chosen. 

Let us consider the following example : 

A paraboloid of revolution has its axis coincident with the 
diameter of a sphere, and its vertex in the surface of the sphere ; 
required the volume between the two surfaces. 


Let y? = 2 mx Ct) 
be the parabola, and) 2+ 7? —2ax=0 (2) 
be the circle, which form the paraholoid and the sphere by their 
revolution. The abscissas of their points of intersection are 0 
and 2(a—m). 

We have [ee f i ydydx, 
and, in performing our first integration, our limits must be the 
values of y obtained from equations (1) and (2). 


We get k= { [2(a — m)x — 2 |da, 
and here our limits of integration are 0 and 2(a —m). 
] 3 
Hence V=$r(a—m)=", 


if h is the altitude of the solid in question. 


EXAMPLES. 


(1) A cone of revolution and a paraboloid of revolution have 
the same vertex and the same base; required the volume be- 


2 
tween them. = ngs, ae , where / is the altitude of the cone. 





174 INTEGRAL CALCULUS. (ArT. 152. 


(2) Find the volume included between a right cone, whose 
vertical angle is 30°, and a sphere of given radius touching it 
along a circle. pls oi 


Solids of Revolution. Polar Formula. 


152. If we use polar codrdinates, and suppose the revolving 
area broken up, as.in Art. 188, into elements of which rdddr 
is the one at any point P whose codrdinates are r and ¢, the 
element rddédr will generate a ring whose volume will differ 
from 277° sin dédd¢dr by an infinitesimal of higher order than the 
second, if we regard d¢ and dr as of the first order; for it will 
be less than a prism having for its base rd¢édr, and for its alti- 
tude 27(r+dr) sin(¢ +d¢), and greater than a prism having 
the same base and the altitude 277rsin¢d; and these prisms 
differ by an amount which is infinitesimal of higher order than 
the second. 

We shall have then 


V=2 rf {7 sin ¢ddrd¢, [1] 
the limits being so taken as to bring in the whole of the gener- 
ating area. 

For example ; let us find the volume generated by the revolu- 
tion of a cardioide about its axis. 


r= 2a(1—cos¢) 
is the equation of the cardioide ; 


Vie 2x ff rsin dards. 


Our first integral must be taken between the limits 7 = 0 and 
r= 2a(1—cos¢), and is 


ra — cos ¢)’sin ddd. 


ead Wes Gui maps re 
V= at fa cos d)* sin ddd, 


V= sna, 


Cuap. XII.] VOLUMES. 175 


EXAMPLE. 


A right cone has its vertex on the surface of a sphere, and its 
axis coincident with the diameter of the sphere passing through 
that point; find the volume common to the cone and the sphere. 


Volume of any Solid. Triple Integration. 


153. If we suppose our solid divided into parallelopipeds by 
planes parallel to the three codrdinate planes, the elementary 





parallelopiped at any point (x,y,z) within the solid will have for 
its volume AvAyAz, or, if we regard a, y, and 2 as independent, 
dadydz; and the whole volume 


V=f ff dxdyae, 1] 


the limits being so chosen as to embrace the whole solid. 

The integrations are independent, and may be performed in 
any order if the limits are suitably chosen. 

As it is important to have a perfectly clear conception of the 
geometrical interpretation of each step in the process of finding 


176 INTEGRAL CALCULUS. TART. 153. 
4 


a volume by a triple integration, we will consider one case in 
detail. 
Let the integrations be performed in the order indicated by 


the formula 
Yes ij} { if dedyda. 


If the limits are correctly chosen, our first integration gives 
us the volume of a prism one of whose lateral edges passes 
through any chosen point P,(2,y,z) within the solid, is parallel 
to the axis of Z, and reaches directly across the solid from 
surface to surface, while the base of the prism is the rectangle 
dydx; our second integration gives the volume of a right cylin- 
der whose base is a plane section of the solid, passes through 
the point P, and is parallel to the plane YZ, and whose altitude 
is dx; and our third integration gives the volume of the whole 
solid. | 

The limits in our first integration are, then, the values of z 
belonging to the point in the lower bounding surface and the 
point in the upper bounding surface which have the codrdinates 
x and y; the limits in the second integration are the values of y 
belonging to the two points in the perimeter of the projection 
of the solid in the plane of X Y which have the codrdinate 2; 
and the limits in the third integration are the least value and 
the greatest value of « belonging to points on the perimeter of 
the projection of the solid on the plane of XY. 

It is easily seen from what has just been said that the limits 
in the second and third integrations are precisely those we 
should use if we were finding the area of the projection of the 


solid by the fornfula 
A ih f dydx. 


Of course, it is necessary to have a clear idea of the form of 
the solid whose volume is required. 
For example, let us find the volume of the portion of the 
ellipsoid a? 2 2 
Of tase 
ls Ae ee | 
a? am ee Migs 


cut off by the codrdinate planes. 


Cuap. XII.] VOLUMES. LEG 


V= ) f fdedyac, 


) 


Le, 

r Oat il: 

b4| i-=,; and for x, 0 and a. For, starting at any point 
al 


| 
and our limits are, for z, 0 and c4/1— ; for y, 0 and 





(x,y,z) and integrating on the hypothesis that z alone varies, we 
get a column of our elementary parallelopipeds having dady as a 
base and passing through the point (7,y,z). To make this col- 
umn reach from the plane X Y to the surface, z must increase 
from the value zero to the value belonging to the point on the 
surface of the ellipsoid which has the codrdinates x and y; that 
is, to the value eyi-—= ee Then integrating on the hy- 
fea bee 

pothesis that y alone varies, we shall sum these columns and 
shall get a slice of the solid passing through (7,y,z) and having 
the thickness dx. T’o make this slice reach completely across 
the solid, we must let y increase from the value zero to the 
greatest value it can have in the slice in question; that is, to the 
- value which is the ordinate of that point of the section of the 
ellipsoid by the plane X Y which has the abscissax. The section 
in question has the equation 





ne 
therefore the required value of y is } \ Tits a 
a 


Last, in integrating on the hypothesis that x alone varies, we 
must choose our limits so as to include all the slices just cle- 
scribed, and must increase x from zero to a. 


a 2 
faeseneyi-%-¥ 


between the limits 0 and 





178 INTEGRAL CALCULUS. [ARr. 153. 








the volume required. 


EXAMPLES. 


(1) Find the volume obtained in the present article, perform- 
ing the integrations in the order indicated by the formula, 


V fave adydz. 


(2) Find the volume cut off from the surface 


2 2 
Cc b 


by a plane parallel to that of YZ, at a distance a from it. 
Ans. ra?V (bc). 


(3) Find the volume enclosed by the surfaces, 





of +e a? = 02,07 ig" = ane eee Mie 3 rat 
32¢ 
(4) Obtain the volume bounded by the surface 
z=a—Vae+y? 
and the planes Co 2 Andy Biee0. Ans. ag. 


Cnap. XII.} VOLUMES. 179 


(5) Find the volume of the conoid bounded by the surface 
rea 





Ss age 
z+ —+-=c’ and the planes x=0 and x=a. Ans. 


a 


154. If we use polar codrdinates we can take as our element 


of volume 
7” sin ddrdddé, 


an expression easily obtained from the element 277? sin¢drdd 
used in Art. 152. 


Then ies f f f ? sin ddrdddd, 


where the order of the integrations is usually immaterial if the 
limits are properly chosen. 
EXAMPLES. 
(1) Find the volume of a sphere by polar codrdinates. 
(2) Find the whole volume of the solid bounded by 
(P+ yt 2) = 27 aS xyz. 


Suggestion: ‘Transform to polar codrdinates. Ans. ~a’, 


180 INTEGRAL CALCULUS. [ART. 155. 


CH ADE KR Xeni 
CENTRES OF GRAVITY, 


155. The moment of a force about an axis perpendicular to its 
line of direction is the product of the magnitude of the force by 
the perpendicular distance of its line of direction from the axis, 
and measures the tendency of the force to produce rotation 
about the axis. 

The force exerted by gravity on any material body is propor- 
tional to the mass of the body, and may be measured by the 
mass of the body. 

The Centre of Gravity of a body is a point so situated that the 
force of gravity produces no tendency in the body to rotateabout 
any axis passing through this point: 

The subject of centres of gravity belongs to Mechanics, and 
we shall accept the definitions and principles just stated as data 
for mathematical work, without investigating the mechanical 
grounds on which they rest. 


156. Suppose the points of a body referred to a set of three 
rectangular axes fixed in the body, and let %,7,% be the codrdi- 
nates of the centre of gravity. Place 
the body with the axes of X and Z 
horizontal, and consider the tendency 
of the particles of the body to produce 
rotation about an axis through (2%, 7,2) 
parallel to OZ, under the influence of 
oravity. Represent the mass of an 
elementary parallelopiped at any point 
(x,y,z) by dm. The force exerted by 
gravity on dm is measured by dm, and 
its line of direction is vertical. If the mass of dm were concen- 
trated at P, the moment of the force exerted on dm about the 





Cuap. XIII.] CENTRES OF GRAVITY. 181 


axis through C would be («—%)dm, and this moment would 
represent the tendency of dm to rotate about the axis in ques- 
tion ; the tendency of the whole body to rotate about this axis 
would be 3(@—2)dm. If now we decrease dm indefinitely, the 
error committed in assuming that the mass of dm is concentrated 
at. P decreases indefinitely, and we shall have as the true expres- 
sion for the tendency of the whole body to rotate about the axis 


through O, 4) («—x)dm; but this must be zero. 


Hence foe —x)dm=0, 


if adn — % if ‘dm tt 
iff adm 

[1] 
aft dm 


If we place the body so that the axes of Y and X are hori- 
zontal, the same reasoning will give us 


ff. ydm £2] 





¢= 








z= — . [3] 
Since | dm is the mass of the whole body, if we represent it 


by M@ we shall have 
fcam 





b) 








182 INTEGRAL CALCULUS. [Arr. 157. 


EXAMPLE. 


Show that the effect of gravity in making a body tend to rotate 
about any given axis is precisely the same as if the mass of the 
body were concentrated at its centre of gravity. 


157. The mass of any homogeneous body is the product of 
its volume by its density. If the body is not homogeneous, the 
density at any point will be a function of the position of that 
point. Let us represent it by x. Then we may regard dm as 
equal to «dv if dv is the element of volume, and we shall have 





“fae | Ut] 


and corresponding formulas for 7 and 2. 
If the body considered is homogeneous, « is constant, and we 
shall have 








4 9 2 
fw ye L : 


as fav Vv . [3] 


{ 2av { 2av 
Z eae 4 > Seah Sires 


J dw U oe 











In any particular problem we have only to express dv in 
terms of the coordinates. | 
Plane Area. 


158. If we use rectangular codrdinates, and are dealing with 
a plane area, where the weight is uniformly distributed, we have 


dv =dA = dady. (Art. 136). 





Cuap. XIII.] CENTRES OF GRAVITY. 183 


Hence, by 157, [2] and [3], 


mal J feaedy xdady 
"of faeay deedy 
ee f Sydea, Westy: | 


| f favay dxdy 


If we use polar codrdinates. 


[1] 


dv=dA=rdd¢ddr, 


so me cos U3 i ouameade | 
; ff fre f rdodr | 


i‘ iG sin $ ddr | 


na [fragar ddr 


For example ; let us find the centre of gravity of the area be- 


tween the cissoid and its asymptote. From the equation of the 
cissoid 


[2] 


iors 


a 
y= ; 
is 





we see that the curve is symmetrical with respect to the axis 
of X, passes through the origin, and has the line ®=a as an 
asymptote. From the symmetry of the area in question, 7 = 0, 
and we need only find 2. 


LS ‘fe adyda nk ay die 
eed OUNCES ae 0 0 

=, 

aaa { fayax dyda H yde 
- 0 





he 


184 INTEGRAL CALCULUS. [Arr. 158. 


6 


Seater pu (ae 
pa eo) ne 


Cae 2 ap Rng ib 
N earmeee da Nesey 
0(a— x) 0(a— a) 


Wa! 
C= Fas 


As an example of the use of the polar formulas [2], let us find 
the centre of gravity of the cardioide 
r= 2a(1—cos¢). 


Here, from the fact that the axis of X is an axis of symmetry, 


we know that 7 = 0. 


a | Seos andes cos ddrdd 
ye ae raragp 
af cos ies Sy 


ate dd eo 


(cose — 8 cos +3 cos*d— cos*d)dd =— 1b 7; 


c= 


cos se 

















and fa — 2cos¢ + cos’ )dd = 37. 
0 


Hence e=— ha. 
EXAMPLES. 


1. Show that formulas [1] hold even when we use oblique 
coordinates. 
2. Find the centre of gravity of a segment of a parabola cut 


off by any chord. 
Ans. «=2a, y=0. If the axes are the tangent parallel 


to the chord and the diameter bisecting the chord. 


Cap. XIII.] CENTRES OF GRAVITY. | 185 


3. Find the centre of gravity of the area bounded by the semi- 
cubical parabola ay’? = x* and a double ordinate. Ans. «= 22. 


4. Find the centre of gravity of a semi-ellipse, the bisecting 
line being any diameter. 


Ans. If the bisecting diameter is taken as the axis of Y, and 


the conjugate diameter as the axis of X, % = : “, y=0. 





ry 


OT 


ol — 2 





5. Find the centre of gravity of the curve y° =) 


ANS. So 2G. 

6. Find the centre of gravity of the cycloid. 
ANS.n& = ar, YozRa. 
7. Find the centre of gravity of the lemniscate r? = «? cos2 ¢. 


7 





x 2 
ANS) Wie a. 
8 

8. Find the centre of gravity of a circular sector. 


Ans. If we take the radius bisecting the sector as the axis 
_2,asia 
= 3 : 





of X, and represent the angle of the sector by 2a, 
a 


9. Find the centre of gravity of the segment of an ellipse cut 








off by a quadrantal chord. Ans. = 4 
10. Find the centre of gravity of a quadrant of the area of the 


curve a+ y3= ai. Ans. ®=y = 2 


159. If we are dealing with a homogeneous solid formed by 
the revolution of a plane curve about the axis of X, we have 


dv = 2rydydx. (Art. 151 [1]) 
Hence, by Art. 157 [2], 
= ale Eu] 


2 
{ { ydaedy 


186 INTEGRAL CALCULUS. [ART. 159. 
If we use polar codrdinates, 

dy = 2rr* sin¢drdd. (Art. 152 [1].) 

(ie sin ¢ cos ddrdd | 


ih fe sin ddrd¢d 


For example ; let us find the centre of gravity of a hemisphere. 
The equation of the revolving curve is 2? + y? == w’. 


Hence “= 


[2] 


a Xar—x 


avydyd 
I, foudy as 


> sp 
v= eee = = fl. 
a Ya? — 2? 4a 


ydydx 
f a, ydy 


If we use polar codrdinates the equation of the revolving curve 
Giga 





ot ng 
oO Oy ~ 
f fo sing cos¢d¢dr |, 
ee 0 0 g ft 


a at 4 ae 
afi 4f; ” sin ddddr 
0 0 


EXAMPLES. 


Here 





‘in OEE 
= ga. 


wi 


1. Find the centre of gravity of the solid formed by the revolu- 
tion of the sector of a circle about one of its extreme radii. 


Ans. = #acos’*4B, where B is the angle of the sector. 


2. Find the centre of gravity of the segment of a paraboloid 
of revolution cut off by a plane perpendicular to the axis. 


Ans. ©=%a, where x =a is the plane. 


3. Find the centre of gravity of the solid formed by scooping 
out a cone from a given paraboloid of revolution, the bases of 
the two volumes being coincident as well as their vertices. 

Ans. The centre of gravity bisects the axis. 


Cuap. XIII. ] CENTRES OF GRAVITY. 187 


4. A cardioide is made to revolve about its axis; find the 
centre of gravity of the solid generated. Ans. F=— 8a. 


5. Obtain formulas for the centre of gravity of any homo- 
geneous solid. 


6. Find the centre of gravity of the solid bounded by the 
surface z? = wy and the five planes x=0, y=0, z=0, wv=a, y=b. 
Ans. =a, Y= 3), Z= Sardi. 


160. If we are dealing with the arc of a plane curve, the 
formulas of Art. 157 reduce to 





[1] 





EXAMPLES. 


1. Find the centre of gravity of an arc of a circle, taking the 
diameter bisecting the arc as the axis of X and the centre as the 


cr pet ae : 
origin. Ans. © = —, where ¢ is the chord of the arc. 
s 


2. Find the centre of gravity of the arc of the curve x§+yi=«i 
between two successive cusps. ANsi 2 AY = 4 4; 


3. Find the centre of gravity of the arc of a semi-cycloid. | 
Ans. =(r—4)a, y=—da. 
4. Find the centre of gravity of the arc of a catenary cut off 
by any horizontal chord. 
Ans. 7=0, ¥= i a *Y, where 2s is the length of the are. 
2s 


=_ 





5. Obtain formulas for the centre of gravity of a surface of 
revolution, the weight being uniformly distributed over the 
surface. 


188 INTEGRAL CALCULUS. (Arr. 161. 


6. Find the centre of gravity of any zone of a sphere. 
Ans. The centre of gravity bisects the line joining the centres 
of the bases of the zone. 


7. A eardioide revolves about its axis; find the centre of 
gravity of the surface generated. Ans. x«=—102a,. 


8. Find the centre of gravity of the surface of a hemisphere 
when the density at each point of the surface varies as its per- 
pendicular distance from the base of the hemisphere. 

Ans. Be 4a, 


9. Find the centre of gravity of a quadrant of a circle, the 
density at any point of which varies as the nth power of its 
distance from the centre. Ans, ieee ee 220. 

n+3 7 

10. Find the centre of gravity of a hemisphere, the density 

of which varies as the distance from the centre of the sphere. 
Ans. = 24. 


Properties of Guldin. 


161. I. If a plane area revolve about an axis external to 
itself through any assigned angle, the volume of the solid gene- 
rated will be equal to a prism whose base is the revolving area 
and whose altitude is the length of the path described by the 
centre of gravity of the area. 


Il. If the arc of a plane curve revolve about an external axis 
in its own plane through any assigned angle, the area of the 
surface generated will be equal to that of a rectangle, one side 
of which 1s the length of the revolving curve, and the other the 
length of the path described by its centre of gravity. 

First; let the area in question revolve about the axis of X 
through an angle ©. The ordinate of the centre of gravity of 


the area in question is 
fi f ydaudy 
—— by Art. 158 [1]. 


Cuap. XIII.] CENTRES OF GRAVITY. 189 


The length of the path described by the centre of gravity 


hey ame (1) 
The volume generated is 


y= of f ydady, by Art. 151. 


Hence V=y0 Bb fi dady. 


But f f dady is the revolving area, and the first theorem is 


established. 
We leave the proof of the second theorem to the student. 


EXAMPLES. 


1. Find the surface and volume of a sphere, regarding it as 
generated by the revolution of a semicircle. 


2. Find the surface and volume of the solid generated by the 
revolution of a cycloid about its base. 


3. Find the volume and the surface of the ring generated by 
the revolution of a circle about an external axis. 

Ans. V=27'a*b, S=47'ab, where b is the distance of 
the centre of the circle from the axis. 


4, Find the volume of the ring generated by the revolution of 
an ellipse about an external axis. 

Ans. V=27°ubc, where c is the distance of the centre of the 
ellipse from the axis. 


190 INTEGRAL CALCULUS. (ArT. 162. 


CHAPTER XIV. 
LINE, SURFACE, AND SPACE INTEGRALS. 


162. Any variable which depends for its value solely upon 
the position of a point, as, for example, any function of the 
rectangular or polar codrdinates of the point, may be called 
a point-function. 

A point-function is said to be continuous along a given line 
if its value changes continuously as the point, on whose position 
the function depends for its value, moves along the line; it is 
said to be continuous over a given surface if its value changes 
continuously as the point is made to move at pleasure over the 
surface; and it is said to be continuous throughout a given 
space if its value changes continuously as the point is made to 
move about at pleasure within the space. 


163. If a given line is divided in any way into infinitesimal 
elements, and the length of each element is multiplied by the 
value a given point-function, which is continuous along the line, 
has at some point within the element, the limit approached by 
the sum of these products as each element is indefinitely de- 
creased, is called the line integral of the given function along 
the line in question. 

If a given surface is divided in any way into infinitesimal 
elements such that the distance between the two most widely 
separated points within each element is infinitesimal, and the 
area of each element is multiplied by the value a given point- 
function, which is continuous over the surface, has at some 
point within the element, the limit approached by the sum of 
these products as each element is indefinitely decreased, is 
called the surface integral of the given function over the surface 
in question. 


CHap. XIV.] LINE, SURFACE, SPACE INTEGRALS. co 


If a given space is divided in any way into infinitesimal 
elements such that the distance between the two most widely 
separated points within each element is infinitesimal, and the 
volume of each element is multiplied by the value a given point- 
function, which is continuous throughout the space, has at 
some point within the element, the limit approached by the 
sum of these products as each element is indefinitely decreased, 
is called the space integral of the given function throughout 
the space in question. 

It is easily seen that the line integral of unity along a given 
line is the length of the line; that the surface integral of unity 
over a given surface is the area of the surface; and that the 
space .ntegral of unity throughout a given space is the volume 
of the space. 

In the chapter on Centres of Gravity we have had numerous 
simple examples of line, surface, and space integrals. 


164. That the value of a line, surface, or space integral is 
independent of the position in each element of the point at 
which the value of the given function is taken can be proved 
as follows: The distance apart of any two points in the same 
infinitesimal element is infinitesimal (Art. 163), therefore the 
values of a continuous function taken at any two points in 
the same element will differ in general by an infinitesimal; the 
products obtained by multiplying these two values by the mag- 
nitude of the element will, then, differ by an infinitesimal of 
higher order than that of the element; therefore, in forming 
the integral either of these products may be used in place of 
the other without changing the result. (I. Art. 161.) 


165. The line integral of a function along a given line is 
absolutely independent of the manner in which the line is 
broken up into infinitesimal elements, and is equal to the length 
of the line multiplied by the mean value of the function along 
the line; the mean value of the function being defined as fol- 
lows: Suppose a set of points uniformly distributed along the 


192 INTEGRAL CALCULUS. [ART. 165. 


line, that is, so distributed that the number of points in any 
portion of the line is proportional to the length of the portion ; 
take the value of the function at each of these points; divide 
the sum of these values by the number of the points; and the 
limit approached by this quotient as the number of the points 
is indefinitely increased is the mean value of the given function 
along the line; and this mean value is in general finite and 
determinate. 

To prove our proposition, we have only to consider in detail 
the method of finding the mean value in question. Let the 
number of points in a unit of length of the line be &. Then, 
no matter how the line is broken up into infinitesimal elements, 
the number of points in each element is & times the length of the 
element. Since any two values of the function corresponding to 
points in the same element differ by an infinitesimal, in finding 
our limit we may replace all values corresponding to points in 
the same element by any one; hence the sum of the values cor- 
responding to points in the same element may be replaced by one 
value multiplied by the number of points taken in that element, 
that is, this sum may be replaced by & times the product of one 
value by the length of the element; and the sum of the values 
corresponding to all the points taken in the line may be replaced 
by & times the sum of the terms obtained by multiplying the 
length of each element by the value of the function at some 
point within the element. When we divide this sum by the whole 
number of points considered, that is, by & times the length of 
the line, the A’s cancel out, and the required mean value reduces 
to the limit of the numerator divided by the length of the line, 
and the limit of the numerator is the line integral of the func- 
tion along the line. Therefore the line integral is the mean 
value of the function multiplied by the length of the line. 

The same proof may be given for a surface integral or for a 
space integral. The former is the product of the area of the 
surface by the mean value of the function over the surface ; 
the latter is the volume of the space multiplied by the mean 
value of the function throughout the space; and both are inde- 


Cap. XIV.] LINE, SURFACE, SPACE INTEGRALS. 193 


pendent of the way in which the surface or space may be divided 
into infinitesimal elements. 


166. If the line along which the integral is taken is a plane 
curve, it is easy to get a geometrical representation of the 
integral. For, if at every point of the line a perpendicular to 
the plane of the line is erected whose length is equal to the 
value of the function at the point, the line integral required 
clearly represents the area of the cylindrical surface containing 
the perpendiculars if the values are all of the same sign, and 
represents the difference of the areas of the portions of the 
cylindrical surface which lie on opposite sides of the line if the 
values of the function are not all of the same sign. 

A similar construction shows that a surface integral over a 
plane surface may be represented by a volume or by the differ- 
ences of volumes. Consequently, in each case if the function 
is finite and continuous, the integral is finite and determinate. 


167. As examples of line, surface, and space integrals, we 
will calculate a few moments of inertia. 

The moment of inertia of a body about a given axis may be 
defined as the space integral of the product of the density at 
any point of the body by the distance of the point from the 
axis; the integral being taken throughout the space occupied 
by the body. 

If the body considered is a material surface or a material 
line, the integral reduces to a surface integral or to a line 
integral. 

In the examples taken below the body is supposed to be 
homogeneous. 


(a) The moment of inertia of a circumference about a given 
diameter. 
Using polar codrdinates and taking the diameter as our axis, 


2r 
I ={ a cos’ ¢.kadd = ka’ r 
0 
=} Mo’, 7 fila 


194 INTEGRAL CALCULUS. (ArT. 167. 


if J is the moment of inertia, and @ the radius, & the density, 
and M the mass of the circumference in question. 


(6) The moment of inertia of the perimeter of a square about 
an axis passing through the centre of the square and parallel 
to a side. 


Ten { “ytkdy +2 uf “a2 eda 


= ika’+4ka' =1%ka’ 
— 3 Ma’, ) [2] 
if 2a is the length of a side. 
(c) The moment of inertia of a circle about a diameter. 
ie A ; fe sin’ . krdbdr = dhenat 
=} Ma’. [3] 


(a) The moment of inertia of a square about an axis through 
the centre of the square and parallel to a side. 


T= {_ f yhdedy = {ha 
= } Ma’. | [4] 


(ec) The moment of inertia of the surface of a sphere about 
a diameter. 


yee f 4 a? sin? . ka? sin pdodé = Skwat! 
0 WJ0 
= 2 Ma’. [5] 


(f) The moment of inertia of the surface of a cube about an 
axis parallel to an edge and passing through the centre. 


ren 49" [(@+eykdade +2 f° f+ e)kdyde 


= 10 Ma’. [6] 





Cap. XIV.] LINE, SURFACE, SPACE INTEGRALS. 195 


(g) The moment of inertia of a sphere about a diameter. 


20 T a 
ne ap if of; 2 sin? . br? sin o drdodé = 8 kma? 
0 0 cit 
— 2 Mo’. [7] 


(h) The moment of inertia of a cube about an axis through 
the centre and parallel to an edge. 


i As [G+ #)kdadyd: = 18 ka? 


= 3 Ma’. [8] 
EXAMPLES. 


Find the moments of inertia of the following bodies: 


(1) Of a straight line about a perpendicular through an 
extremity ; about a perpendicular through its middle point. 

(2) Of the circumference of a circle about an axis through 
its centre perpendicular to its plane. Ans. Ma’. 


(5) Of a circle about an axis through its centre perpendicular 
to its plane. Ans. 4 Ma’. 


(4) Of a rectangle whose sides are 2a, 2b, about an axis 
through its centre perpendicular to its plane; about an axis 
through its centre parallel to the side 20. 

Ans. $M(a? + 6”); 4 Ma’. 

(5) Of an ellipse about its major axis; about its minor axis ; 
about an axis through the centre perpendicular to the plane of 
the ellipse. Ans. Mb’; + Ma’; 4 M(a’? +b’) 


(6) Of an ellipsoid about the axis a. Ans. 1M(b? +c’). 


(7) Of a rectangular parallelopiped about an axis through 
the centre parallel to the edge 2a. Ans. 1 M(b’? +’). 


(8) Of a segment of a parabola about the principal axis. 
Ans. ?Mb’, where 26 is the breadth of the segment. 


196 INTEGRAL CALCULUS. [ART. 168. 


168. If u, D,u, and D,u are finite, continuous, and single- 
valued for all points in a given plane surface bounded by a 
closed curve 'T, the surface integral of Du taken over the surface 
is equal to the line integral of ucosa taken around the whole 
bounding curve, where a is the angle made with the axis of X 
by the external normal at any point of the boundary. 


This may be formulated thus: 


ff. udady = fv cosa. ds. [1] 


Let the axes be chosen so that the surface in question lies in 
the first quadrant, and divide the projection of 7 on the axis 
of Y into infinitesimal elements of which any one is dy. 





On each of these elements as a base erect a rectangle; and 
since 7’ is a closed curve, each of these rectangles will cut it 
an even number of times. 

Let us call the values of w at the points where the lower side 
of any one of these rectangles cuts 7, w%, Us, Ug, U4, ete., re- 
spectively ; the angles which this side makes with the exterior 
normals at these points, a), a, a3, a4, etc.; and the elements 
which the rectangle cuts from 7’, ds,, ds., ds3, ds,, ete. 

It is evident that whenever a line parallel to the axis of X 
cuts into the surface bounded by T, the corresponding value of 
x is obtuse and its cosine negative ; that whenever it cuts out, 


Cuap. XIV.] LINE, SURFACE, SPACE INTEGRALS. 197 


a is acute and its cosine positive; and that any value of a is 
the angle which the contour 7’ itself makes at the point in ques- 
tion with the axis of Y if we suppose the contour traced by a 
point moving so as to keep the bounded surface always on the 
left hand. 

We have then approximately, 


dy= — ds, + COS 0;= ds, + COS ay= — 8; + COSaz=ds,+ COSay=-++. [2] 


If, now, in if uf: D,udxdy we perform the integration with 


respect to 2, and introduce the proper limits, we shall have 


ff? udady = { ay( — Uy + Ug — Us + Uy:++)5 [3] 


and the second member indicates that we are to form a quantity 
corresponding to that in parentheses for every rectangle which 
cuts 7’, to multiply it by the base of the rectangle, and then to 
take the limit of the sum of the results as all the bases are 
indefinitely decreased. f 


By [2], | 
dy (— Uy 4- Ug — Uz + Uy?) 
= U; COS a; AS, + Uy COS ay U8, + Uz COS ag sz + U,COSa,ds,-+ +--+; [4] 


and the limit of the sum of the values any one of which is 
represented by the second member of [4] is clearly | «cosads 


taken around the whole of T. 


EXAMPLE. 


Prove that under the conditions stated in the last article 


f D,udady = | ucosB.ds, 


where £ is the angle made with the axis of Y by the exterior 
normal. 


INTEGRAL CALCULUS. [ArT. 169. 


198 
169. As an illustration of the last proposition, let us find the 


centre of gravity of a semicircle. 
(1) 


Eig 
= i dady. 


We have 
Hence, by Art. 168, 


But we may write y= D,(xy). 
a= = ff yaeay = = fz cos ads 
k x a pe 2 
— oO) acos¢dasing cosd¢add +f. re We cos - de 4 
0 —a 


2 Say et oe 


7 28 or. 








which agrees with the result of Ex. 8, Art. 158. 
As a second example, we shall find the moment of inertia of 
a circle about a diameter. 
We have 
I= kf yPdedy =k f ay cos ¢. ds 
= kf cosda’ sin’ d cos dad 
0 
ary t vf 2 2 k 4 1 2 
=ka* | sin*¢ cos Sadie chore la ae 
e 0 ; “ 


which agrees with the result of (¢), Art. 6. 
EXAMPLES. 
(1) Find the centre of gravity of a semicircle, using the 


theorem f { Dynacay a fru cos B.ds. 


(2) Find the moment of inertia of a circle about an axis 
through its centre perpendicular to its plane, using the principles 


f | ‘D, udady = ucosa.ds and f _f Dyuderdy = ucosB.ds 


Cuap. XIV.] LINE, SURFACE, SPACE INTEGRALS. 199 


170. Since, as we have seen in Art. 168, a is the angle which 
the curve 7’ makes with the axis of Y; if we trace the curve 
so as to keep the bounded space on our left, it follows that 
cosa.ds = dy. 


Hence ff2. udady = | udy ; [1] 


and in like manner, 


{_{ Dyudeay a — {ude : 2] 


the first integral in [1] and [2] being taken over the bounded 
surface, and the second around the bounding curve. 

For example, the moment of inertia of a square about an 
axis through the centre and parallel to a side is 


jue kf { yPdady. ((d) Art. 167.) 


By [1], f fyeady = fray, 


and the last integral is to be taken around the perimeter. 
Hence 


hee K| faotay +f (= ayy) | =e haf ydy = $ka* 


— $ Ma?. 
EXAMPLE. 


Work Ex. 7, Art. 167, by the aid of (2). 


171. ff U, D,U, D,U, and D,U are finite, continuous, 
single-valued functions throughout the space bounded by a given 
closed surface T, the space integral of D,U taken throughout the 
space in question is equal to the surface integral, taken over the 
bounding surface, of Ucosa, where a is the angle made with 
the axis of X by the exterior normal at any point of the surface. 

This may be formulated thus : 


ff J B.Warayaz= ('Veosa.ds. [1] 


200 INTEGRAL CALCULUS. (Art. 171. 


nl 


The proof is almost identical with that given in Art. 168, 
except that for elementary rectangle we use elementary prism. 


We shall merely indicate the steps. 


Y 

















~oe 0 


Z 


dydz = — dS, cosa, = dS, cosa, = — AS; cosa; = «+. 


ff (_D, Udedydz ={ {we[-U4+0-U,-] 


= the limit of the sum of terms of the form 


U, cosa,. dS, + Uy, cosa. dS, + U3 cosag.dSz + ++ 


= { Ucosa dS. 


EXAMPLE. 


Prove that under the conditions of the last article 


ik { afin, Udadydz ={u cosB. dS, 
and { 1p 2. Udadydz = iff Ucosy.ds, 


where 6 and y-are the angles made with the axes of Y and Z 
respectively by the exterior normal to the bounding surface. 


Cuap. XIV.) 


LINE, SURFACE, SPACE INTEGRALS. 201 


172. As an illustration, let us find the centre of gravity of a 


hemisphere. 


We have 


= = ff xdxdyaz = [= cosa.ds 





27 ‘ 
se ch “ a cos’ cos pa’ sind dddé 








Wore 
ey Bite 


which agrees with the result of Art. 159. 


EXAMPLE. 


Find the moment of inertia of a sphere about a diameter; of 
a cube about an axis through the centre parallel to an edge. 
Make your work depend upon finding the value of a surface 


integral. 


202 INTEGRAL CALCULUS. (ART. 173. 


CHAPTER XV. 
MEAN VALUE AND PROBABILITY. 


173. The application of the Integral Calculus to questions 
in Mean Value and Probability is a matter of decided interest ; 
but lack of space will prevent our doing more than solving 
a few problems in illustration of some of the simplest of the 
methods and devices ordinarily employed. A full and admirable 
treatment of the subject is given in ‘* Williamson’s Integral 
Calculus” (London: Longmans, Green, & €o.); and numer- 
ous interesting problems are published with their solutions 
in ‘*The Mathematical Visitor” and ‘* The Annals of Mathe- 
matics.” 


174. The mean of n quantities is their sum divided by their 
number. If the number of quantities considered is supposed 
to increase indefinitely according to some given law, the prob- 
lem of finding the limiting value approached by their mean 
usually calls for the Integral Calculus. The mean value of a 
continuous function of one, two, or three independent variables 
has been carefully defined in Art. 165, and has been proved to 
depend upon a line, surface, or space integral. 


(a) Let us find the mean distance of all the points on the 
circumference of a circle from a given point on the circumfer- 
ence. 

If we take the given point as origin, the distances whose 
mean is required are the radii vectores of points uniformly dis- 
tributed along the circumference of the circle. 

The required mean is, therefore, by Art. 165, equal to 


Cuap. XV.] MEAN VALUE AND PROBABILITY. 203 


the quotient obtained by dividing the line integral of r taken 
around the circumference by the length of the circumference ; 


that is, 
f rds 
M= 





2 3a 
The polar equation of the circle is 


r= 2ac08S¢;} 


ds = 2.adq, 
1 2 
M= — 4 a cos¢dd = 
2 ral 


the required mean value. 


(6) Let us find the mean distance of points on the surface 
of a circle from a fixed point on the circumference. 

Here, by Art. 165, the required mean is the surface integral 
of r taken over the circle, divided by the area of the circle ; 
that is, 
2a COBH 


pe ‘dnd ==". 








(c) The problem of finding the mean distance of points on 
the surface of a square from a corner of the square can be sim- 
plified slightly by considering merely one of the halves into 
which the square is divided by a so Sere 

Here 


asech 


a uf. rdrdd 


al iiZx 37 
=-— 7) loo tan — )- 
(7 + log an=Z) 


204 INTEGRAL CALCULUS. (Arr. 175. 


(d) As an example of a device often employed, we shall now 
solve the problem, To find the mean distance between two points 
within a given circle. 

If M be the required mean, the sum of the whole number of 
cases can be represented by (7r’)’M, 7 being the radius of the 
circle ; since for each position of the first point the number of 
positions of the second point is proportional to the area of the 
circle, and may be measured by that area; and as the number 
of possible positions of the first point may also be measured 
by the area of the circle, the whole number of cases to be con- 
sidered is represented by the square of the area; and the sum 
of all the distances to be considered must be the product of the 
mean distance by the number. 

Let us see what change will be produced in this sum by in- 
creasing 7 by the infinitesimal dr; that is, let us find d(7*r*M). 

If the first point is anywhere on the annulus 2 7z7r.d7, which we 
have just added, its mean distance from the other points of the 





Therefore, the sum of the new distances, to be considered, 


if the first point is on the annulus, is at. 77”. 2rrdr; but the 


sa 
second point may be on the annulus, instead of the first; so that 


to get the sum of all the new cases brought in by increasing 
r by dr, we must double the value just obtained. 





Hence d (n't M ) = 188 rr'dr, 
a 
matM = 128 5 nf: dr = 128 rab, 
0 
128 
Me 
457 


175. In solving questions in Probability, we shall assume 
that the student is familiar with the elements of the theory as 
given in ‘‘ Todhunter’s Algebra.” 


(a) A man starts from the bank of a straight river, and 
walks till noon in a random direction; he then turns and walks 


Cuap. XV.] MEAN VALUE AND PROBABILITY. \< 205 


in another random direction ; what is the probability that he wilt 
reach the river by night? 

Let 6 be the angle his first course makes with the river. If 
the angle through which he turns at noon is less than 7 — 26, 
he will reach the river by night. For any given value of 6, 
z— 70" ‘The probability that 


20 
6 shall lie between any given value 6) and 6) + dé is “. 





then, the required probability is 


The chance that his first course shall make an diate with the 
river between 6) and @) + d6, and that he shall get back, is 


a—20 db _ (77 — 20)d6 


© 





2a 47 7 


As @ is equally likely to have any value between 0 and 5, the 
required probability, 


0 vie 


(6) A floor is ruled with equidistant straight lines; a rod, 
shorter than the distance between the lines, is thrown at ran- 
dom on the floor; to find the chance of its falling on one of the 
lines. 

Let x be the distance of the centre of the rod from the nearest 
line ; 6 the inclination of the rod to a perpendicular to the paral- 
lels passing through the centre of the rod; 2a the common dis- 
tance of the parallels; 2c¢ the length of the rod. 

In order that the rod may cross a line, we must have 
ccos@ > «x; the chance of this for any given value % of 2 is 
1 1% 

C08. —. 
Ao c ae 
The probability that x will have the value x is —. The 
bee ‘ a 
probability required is 
SN } cos de = AG 
TO C 7 

This problem may be solved by another method which pos- 

sesses considerable interest. 


206 INTEGRAL CALCULUS. [ART. 175. 


Since all values of x from 0 to a, and all values of 6 from es 
to 3 are equally probable, the whole number of cases that can 


arise may be represented by 


47 pa 
ih if dxd@ = ra. 


1 
2 


The number of favorable cases will be represented by 


im ccosé 
i dad@ = 2c. 


Hence p=— 


(c) To find the probability that the distance of two stars, 
taken at random in the northern hemisphere, shall exceed 90°. 

Let a be the latitude of the first star. With the star as a 
pole, describe an are of a great circle, dividing the hemisphere 
into two lunes; the probability that the distance of the sec- 
ond star from the first will exceed 90° is the ratio of the lune 
not containing the first star to the hemisphere, and is equal 


to G7=9) The probability that the latitude of the first star 


Me 
will be between a and a+da is the ratio of the area of the 
zone, whose bounding circles have the latitudes a and a+da 
respectively, to the area of the hemisphere, and is 

27a’ cosa da 


= cosa da. 
2 2 
Ta 


Hence ie (Gace) (o7 — 2) cosa da = a 
e’0) T T 


(dq) A random straight line meets a closed convex curve ; 
what is the probability that it will meet a second closed convex 
curve within the first? 

If an infinite number of random lines be drawn in a plane, all 
directions are equally probable; and lines having any given 


Cuap. XV.] MEAN VALUE AND PROBABILITY. 207 


direction will be disposed with equal frequency all over the 
plane. If we determine a line by its distance p from the origin, 
and by the angle a which p makes with the axis of X, we can get 
all the lines to be considered by making p and «a vary between 
suitable limits by equal infinitesimal increments. 

In our problem, the whole number of lines meeting the exter- 


nal curve can be represented by f i) dpda. If the origin is 


within the curve, the limits for p must be zero, and the perpen- 
dicular distance from the origin to a tangent to the curve; and 
for a must be zero and 27. If we call this number N, we 


shall have 
277 
LN) = As pda, 
0 


p being now the perpendicular from the origin to the tangent. 

If we regard the distance from a given point of any closed 
convex curve along the curve to the point of contact of a tan- 
gent, and then along the tangent to the foot of the perpendicu- 
lar let fall upon it from the origin, as a function of the a used 
above, its differential is easily seen to be pda. If we sum these 
differentials from a=0 to a=27, we shall get the perimeter 
of the given curve. 


Hence | N= { pda =) OP 
0 


where ZL is the perimeter of the curve in question. By the same 
reasoning, we can see that », the number of the random lines 
which meet the inner curve, is equal to /, its perimeter. For p, 
the required probability, we shall have 


pas 
: 


EXAMPLES. 


(1) A number x is divided at random into two parts; find the 


mean value of their product. we ne? 


208 INTEGRAL CALCULUS. (ART. 175. 


(2) Find the mean value of the ordinates of a semicircle, sup- 
posing the series of ordinates taken equidistant. Ans. ry 


(3) Find the mean value of the ordinates of a semicircle, sup- 


posing the ordinates drawn through equidistant points on the 
2a 


circumference. 
Ans. 


TT 


(4) Find the mean values of the roots of the quadratic 
« —ax+b=0, the roots being known to be real, but 0 being 
unknown but positive. Ae om ahd 

6 

(5) Prove that the mean of the radii vectores of an ellipse, the 
focus being the origin, is equal to half the minor axis when they 
are drawn at equal angular intervals, and is equal to half the 
major axis when they are drawn so that the abscissas of their 
extremities increase uniformly. 


(6) Suppose a straight line divided at random into three 
parts ; find the mean value of their product. Le a® 


(7) Find the mean square of the distance of a point within a 
given square (side = 2a) from the centre of the square. 

i Ans. 2a’. 

(8) A chord is drawn joining two points taken at random on 

a circumference ; find the mean area of the less of the two seg- 

ments into which it divides the circle. Py ra” a 

T 

(9) Find the mean latitude of all places north of the equator. 

Ans. Ba sas 

(10) Find the mean distance of points within a sphere from 

a given point of the surface. Ans. $a. 


(11) Find the mean distance of two points taken at random 
within a sphere. Ans. 38a. 


(12) Two points are taken at random in a given line a; find 
the chance that their distance shall exceed a given value c. 


2 
Ans. (* a ‘) . 





Cuap. XV.] MEAN VALUE AND PROBABILITY. 209 


(13) Find the chance that the distance of two points within 
a square shall not exceed a side of the square. Ans. r—13. 


(14) A line crosses a circle at random ; find the chances that 

a point, taken at random within the circle, shall be distant from 
the line by less than the radius of the circle. 2 

Ans. 1 — re 


Tv 


(15) A random straight line crosses a circle ; find the chance 
that two points, taken at random in the circle, shall lie on 
opposite sides of the line. . pies 128 

45 x 

(16) A random straight line is drawn across a square ; find 

the chance that it intersects two opposite sides. log 2 








Ans. $— 


(17) Two arrows are sticking in a circular target; find the 
chance that their distance apart. is greater than the radius. 


Ans. * 





VB 
4 


(18) From a point in the circumference of a circular field a 
projectile is thrown at random with a given velocity which is 
such that the diameter of the field is equal to the greatest range 
of the projectile: find the chance of its falling within the field. 


Ans. 4 ae 1). 


(19) On a table a series of equidistant parallel lines is drawn, 
and a cube is thrown at random on the table. Supposing that 
the diagonal of the cube is less than the distance between con- 
secutive straight lines, find the chance that the cube will rest 
without covering any part of the lines. 


Ans. 1 — ae , where a is the edge of the cube and c the dis- 
TT 
tance between consecutive lines. 
(20) A plane area is ruled with equidistant parallel straight 


lines, the distance between consecutive lines being c. A closed 
curve, having no singular points, whose greatest diameter is less 


210 INTEGRAL CALCULUS. (Arr. 175. 


than c, is thrown down on the area. Find the chance that the 
curve falls on one of the lines. 
Ans. oe where / is the perimeter of the curve. 
TC 
(21) During a heavy rain-storm, a circular pond is formed in 
a circular field. If a man undertakes to cross the field in the 
dark, what is the chance that he will walk into the pond? 








Crap. XVI.] ELLIPTIC INTEGRALS. 211 


ipa DE Ry XuVviT. 
ELLIPTIC INTEGRALS. 


: 176. In attempting to solve completely the problem of the 
motion of a simple pendulum by the methods of I. Chapter 
VIII. we encounter an integral of great importance which we 
have not yet considered. ‘The problem is closely analogous to 
that of the Cycloidal pendulum (I. Art. 119). 

For the sake of simplicity we shall suppose the pendulum 
bob to start from the lowest point of its circular path with the 
initial velocity that would be acquired by a particle falling. 
freely in a vacuum through the distance y%; and this by I. Art. 
114 [1] is V2gy. 

Forming our differential equation of motion as in I. Art. 118, 
but taking the positive direction of the axis of Y upward, we 
have Cs ees dy. (1) 
dt? as 


Multiplying by 2° and integrating, 
i 


or, determining C, 


v= ay 29(Yo—y)- (2) 


If the starting-point is taken as the origin, the equation of 
the circular path is 2+ y? —2ay=0, whence 


) = a dy\* 
dt 2ay —y’\dt 


ais -Oetahe d ay ea 
and we have anaes 77 V29(y—Y), 


212 INTEGRAL CALCULUS. [ArT. 176. 


ady 


or dt Res Serge a ee ee 
V2q . Vy —y) Qay— YP) 





Integrating, and determining the arbitrary constant, we get 


pear CON (Tate ets Oot ae (3) 
eek V (yo — y) (2ay — ¥) 


as the time required to reach that point of the path which has 
the ordinate y. 


The substitution of 2? =F reduces (3) to the form 
0 
la . dx 
SA eo eer eee ee 4) 
Noh Sale 
g ore Yo 
2 


where the integral is of the form 








Aer (5) 
0 J —2*)(1— a) 


k? being positive and less than unity if y is less than 2a. An 
examination of equation (2) will show that if this is true, the 
pendulum will oscillate between the two points of the are which 
have the ordinate %. 

If y is greater than 2a, the pendulum will make complete 


revolutions. For this case the substitution of 2 = oe in (3) 
a 


will reduce it to 
y 13 da 
Nyon 2a). (6) 


Va —#)(1 let 


where the integral is of the form (5), k* being positive and less 
than unity. 

The time required for the pendulum to reach its greatest 
height — that is, in the first case, the time of a half-vibration, 
and in the second case, the time of a half-revolution — will 
depend upon 


é da 


0/1) (1 — Katy @) 


Cuap. XVI.] ELLIPTIC INTEGRALS. 213 


177. The length of an arc of an Ellipse, measured from the 
extremity of the minor axis, has been found to be (Art. 107) 


a* — ex" 
oa i) Vest a — @) 
If we replace = by x, (1) becomes 


x — ou-me 
s=a Ne SN lan, (2) 
0 


and the integral is of the form 


f va — kPa? (3) 


where k? is positive and less than unity. 
The length of an Elliptic quadrant depends upon the integral 


i foe 
Ra A 4 
IVES ‘ mao): 


_/ 178. It can be shown by an elaborate investigation, for 
which we have not room, that the integral of any algebraic 
function, which is irrational though containing under the square 
root sign an algebraic polynomial of the third or fourth degree, 
_ can by suitable transformations be made to depend upon one 
or more of the three integrals 











. dx 
ca ae Si ae 1 
pe” Si JG — 2) (1 — #2") iN 
E(k,e) = ri oie . det, 2) 
3 EARN ck ON aE 3 
mati ®) 3 (1 + na?) V (1 — 2) (1 — ha?) i 


which are known as the Elliptic Integrals of the first, second, 
and third class respectively. 


214 INTEGRAL CALCULUS. [Arr. 179. 


k, which may always be taken positive and less than 1, is 
called the modulus; and n, which may be taken real, is called 
the parameter of the integral. 


K= Fk, 1) =( [4] 
0 /(1 — 2°) (1 — #2?) 


py ae 


and Beh, 1) “pba mes dx, [5] 


are known as the Complete Elliptic Integrals of the first and 
second classes. 


179. The substitution of «= sind in the Elliptic Integrals 
reduces them to the following simpler forms. 


TENG aed abc _ (eds 
F(k,¢) ={ J=aan hire [1] . 
E(k, ¢)= ("v1 —esin’d . dd = {"a¢.ag. [2] 
Tees) sea Le oa Oe a eee dg 


> (Ltnsin'd) Vi—-Rsin'e 2 (14nsin’$) Ad 
[3] 


bhai habit -(% og 
» Ji aeane 0 Ag 
Xe 7 

E ={ V1 —k’ sin’ . dd =i Ad .dd. [5] 
0 


@ is called the amplitude of the Elliptic Integral, and ~ 
Ao = V1 —k’sin’¢ is called the delta of ¢, or more simply, 
delta ¢, and is regarded as a new trigonometric function: it is 
always taken with the positive sign, and has an analogy with 
cos d. 

For a given value of k, Ad is easily seen to be a periodic 
function of ¢ having the period 7. It has its maximum value 1 


Crap. XVI] ELLIPTIC INTEGRALS. 215 


when ¢=0 and when ¢=z7, and its minimum value ap (as ies 
which is usually represented by k’ and called the complementary 


modulus, when ¢ = 5 ; and a + «) =A G — «). 


Landen’s Transformation. 


180. The approximate numerical value of an Elliptic Integral 
of the first class, when * and ¢ are given, is easily computed 
by the aid of two valuable reduction formulas due to Landen. 


Tf in F(k, 3) gee. 
——— hee oe Las 


we replace ¢ by ¢,, ¢, and ¢ being connected by the relation 














sin 2 d, 
tan ¢ = ——_, 1 
me k +cos 2 ¢, @) 
which is easily transformable into either of the following: 
k sing = sin (2¢,— ¢), (2) 
bat dite yee oe tang (3) 
v= 4 ek ¢ ly 
py 
fe ee ea . reduces to ; a — { sa eal 
—k* sin’d \! “a : sin’d, 
(1 5 ms 


which is also an Elliptic Integral of the first class, but has a 
different modulus and a Psa amplitude from those of the 
given integral. 

The steps of the process are as follows: 


From (1) we easily find 


HE, As 
ae Os 
1+k?+ 2k cos2 d, 
whence V1 —k’ sin’ d= 1+ cos? ¢, 





V1 +k? + 2k cos2¢, 


216 INTEGRAL CALCULUS. [ART. 180. 


Differentiating (1), we get 























2(1 +k cos2 4) 
sec’ d amas edi : 
a (k + cos2 ¢,)’ ie 
: 1+k°+ 2k cos2 4, 
but f 1 ‘ob = +; 
ut from (1), sec’ (e+ cond aE 
2(1 +k cos 2 ¢;) 
] l@ = —A—____—"_d 
iat ah Wn 1+k?+2k cos2 ¢, Pry 
dd bi 2 dq, ib 2dq, 
V1—Ksint'd VI+H+2kcos2¢, V1+H+2k—4ksin?d, 
2 dq, 
1+k 
cf \ 1— oo sin’, 
(1+k/y 
o, = 0 ~ when cpa IP 
hence yO ae = % fo oe 
o V1—k’sin’d nas Ak 
, lass 2 
NOG 4k (1 Sa 
and F(k, $)= ny 1) | 
4 
where pees 2Vk, [4] 
1+ ke 
and sin(2¢,— ¢)=< sing. 
k, is less than 1 and greater than 4; for otc 1 reduces 
to 0<(1— Vk)?, which is obviously true, and zt Seu 


reduces to 4>k(1++)’, which is true, since & is “0 payin Li. 
If ¢ is not greater than 5 and the smallest value of ¢, con- 

sistent with the relation sin(2¢,—¢)=Xsing is taken, 
0<¢,<¢. Hence (4) is a reduction formula by which we 


can raise the modulus and lower the amplitude of our given 
function. 


Cuap. XVI.] ELLIPTIC INTEGRALS. 217 


By applying the formula (4) n times, we get 








2 2 
Fk; ¢)= —=~ , es ne k ; 
(ios ald Uae ena ergrap ee CO 
or, since a ae tiny) ete., 
1+h Vi 
FG, $)= taf Bates we, 4), 
where os a \ [5] 
k — 2VKiy 1 and sin(2¢,—4¢,_1) =k,_, sin¢,_}. | 
Pp 1+k,4 Pp p Pp P ) 


If we suppose 7 in (9) to be indefinitely increased, we shall 
have pene [k, |= 1; for if we form the series 


ey (hy) (li — ha) >> + (1K) ees, 


we shall have 





| 2N% 
1 — Koya _ +k, Day eh 1—vk, 1 
1—k, 1—k, 1—k; LTR ey 


which is always less than unity ; hence the terms in the series 
must decrease indefinitely as p is increased and hehe [1—k, ]=0. 


Since, as we have seen above, ¢, continually diminishes as 7 
increases, but does not reach the value zero, it must have some 
limiting value @. Hence 


limit edd 
F Kins Pe es F {, ?) = ———— 
[Pay $n)] aye Si /1 — sin? 


n= 
® 
= a) sec ddd = log tan Be ; 
0 ce 


DiwNik hte 
and F(k, ¢) = log tan u aM | yee. [6] 


Formulas [5] and [6] lend themselves very readily to numer- 
ical computation. 


218 INTEGRAL CALCULUS. (ArT. 181. 


181. Formula [4], Art. 180, may be used to decrease the 
modulus and increase the amplitude of a given Elliptic Integral. 
Interchanging the subscripts, and using (3) Art. 180 instead of 
(2) Art. 180, we have 


F(k,¢)=* Ar (hy, $1), | 

where = ere c [1] 
1+V1—-F | 
and tan(¢?,—¢)=V1—F tang. J 


By repeated application of [1] we get 


Fk, b) = "(VRS (eRe + ky) Aw be), | 


Me pe: aes LVL at [2] 
ER Bs RE Di 
and _— tan(¢, — ¢,-1) = Vl — #,_; tang, _- J 


It is easily shown, as in Art. 180, that ee [k, |= 90, and 
N= 0 


limit ® : 
consequently that fiat’ LK, bi) == =f dd =, where ® is the 
0 


limiting value approached by ¢ as n is increased. 


If ‘dame we get from [2], ¢,5=7, ¢d,=27, on hy, == Die 
hence K=F(k, 5)=5a + hy) (1 +h) (1 +its)-+-. [3] 


Formulas [2] and [3], like formulas [5] and [6] of Art. 
180, lend themselves readily to computation. 

With a large modulus, it is generally best to use [5] and [6] 
of Art. 180; with a small modulus, [2] or [3] of the present 
article will generally work more rapidly. 

We give in the next article the whole work of computing the 


Elliptic Integrals, ree 4 by each of the two methods, and 


Cuar. XVI.] ELLIPTIC INTEGRALS. 219 


2 
employing five-place logarithms. 


2 
of computing (Y)- le 4 by the second method, 





182. #3 a Metuop or Arr. 180. 
- = 0.70712 log k = 9.84949 
1-+k = 1.70712 log (1 +k) = 0.23226 
log Vi = 9.92474 
log 2 = 0.30103 
colog (1 + k) = 9.76774 
log ky = 9.99351 
k, = 0.98518 log ky, = 9.99351 


log Vk, = 9.99676 

log 2 = 0.30108 

colog (1 + k,) = 9.70220 
log ky = 9.99999 


Ko == 1 





logk = 9.84949 

log sing = 9.84949 

log sin (24, — ¢) = 9.69898 
2¢,—¢= 30° 0' 3" 
y dee Peat hs pata as te 
Epica. tices inn ae 

log k, = 9.99351 

log sin, = 9.78445 

log sin (2 dy — ¢)) = 9.77796 








220 INTEGRAL CALCULUS. [ART. 182. 


2 do — o, = 36° ob) 3° 
de 74° 21! Be 

Oe idy = 37 oO ees 
5B 63. 35' 16" 


log tan (é 4 10) = 0.30393 
log Vk, = 9.99676 

colog Vk = 0.07526 

log log tan ¢ ti +0) = 9.48277 
colog p = 0.36222 


lox F(“S = = 9.91701 


(2 i= — 0.82605 








x= 0.438429 is the modulus of the common system of 
logarithms. 





ia 


#(%, Ht Metuop or Art. 181. 


VI —k =k! =0.70712 





1—k' = 0.29288 log (1 — k') = 9.46669 
1+k' =1.70712 colog (1 +k!) = 9.76774 
=O. V71Dy log ky = 9.23443 

1 — k, = 0.82843 log (1 — k,) = 9.91826 
1+ k, = 1.17157 log (1 + k,) = 0.06878 
log k,'* = 9.98704 

k,'= 0.98520 logk,' = 9.99352 

1— k,'= 0.01480 log (1 — k,') = 8.17026 
1 + k,'=1.98520 colog (1 + ky')= 9.70220 


k, = 0.00746 log ky = 7.87246 


Cuap. XVI.] | 


1 —k, = 0.99254 
1 + k, = 1.00746 


k.!= 1 
fe = 0 


mrp 


ELLIPTIC INTEGRALS. 


log (1 — k) = 9.99675 
log (1 + ky) = 0.00323 


log k,!? = 9.99998 


logk,’ = 9.99999 


log k' = 9.84949 
log tan ¢ = 0.00000 





log tan (¢, — $) = 9.84949 


$1 — 6 = 35° 15! 53" 


y= 80° 15! 58" 


log k,' = 9.99352 
log tan ¢, = 0.76557 





log tan (d. — ¢;) = 0.75909 


bg ihe Se) B02 7! 


2 = 160° 23' 10" 


tan (o, — do) =tan d, 
® = , = 26, = 320° 46! 20" 


—P= 40° 5! 48" 


= 144348" 
a7 = 648000" 


log ( &\"" = 5.15942 


3 


\- 


colog "= 4.18842 
log r = 0.49715 


1 
log (5°) = 9.84499 


ree 


221 


\ 
222 INTEGRAL CALCULUS. | | AR Trees) 


log (1 + k,) = 0.06878 
log (1 + k,.) = 0.00323 


1 
log (55) = 9.84499 
Mek, 1 = 9.91700 
4 


r(¥, i= — 0.82605 
2° 4 


For r(¥ 4 we have by (3), Art. 181, 


log (1 + k,) = 0.06878 
log (1 + ky.) = 0.00323 
loga = 0.49715 
colog 2 = 9.69897 


log r( 3) = 0.26813 


183. Landen’s Transformation can also be applied to the 
computation of Elliptic Integrals of the second class, but the 
task is a more difficult one; we shall, however, give a brief 
sketch of the method; and in so doing we shall apply it toa 
more general form 

Gh, p)= [SEBS ag, [1] 
V1—k’sin’¢ 
of which £ (k, ¢) is a special case. 
From Art. 180 we have 


- > 1+k 2 
V1 i sin’ — A Oe 








Vi ++ 2k cos2 4, 
Coe 2k + 008 20 aa 
V1 + k?+2kcos2 dq, 

Ns 2(1+ cos2¢y) | 


1+k’?+ 2k cos2 q, 


bo 
bS 
(eu) 


CHap. XVI. ] ELLIPTIC INTEGRALS. 





Hence V1 — sin? +kcos¢ = V1 +h 4+ 2k cos2 dy. 


AM a+ B sin’ d 
G (k, $) =| pa fo 


es a B 1-(1 ape | 
= ee dl 
He te 1—sind * V1—? sin? % 


B 
Wear aa ee an 
=| eel Heng ee: —k* sin b |A, 


and 


G(k, $)—2 sing 


ae 
= 1h aes — 2 (V1 Hsin?d sin + hee) foo 


Substituting ¢, for ¢, this becomes 


a—F cos 24, 


aby: . we ai al ARE ea 
G (k, p) pipet nah 2 f, V1 + k?+2kcos2¢, “os 











Hence G(k, )=Bsi P sing +2 — Gite, [2] 


where 


oe i 2 
ky = ANE, SE ie a SoA a= a— QE, Birra [3] 

Formulas [2] and [3] enable us to make our given function 
depend upon one of the same form, but having a greater 
modulus and a less amplitude. A repeated use of [2], together 
with the reductions employed in Art. 180, gives us 


224 INTEGRAL CALCULUS. [ArT. 183. 


G (k, o)=Fsing + Prsinds + a/2 - By S8in dy 


+E 2. B,8ing;+ -: pee r}B, SIN by_1 





hy es ea | 
Nae (neste CER BD) (4] 
eka ere, 
where By ies ky ko ey 
ly SL EY eran a ae _ 2 
and Ob Seer k apr ae arom Tie eta 1 


Just as in Art. 180 &,, rapidly approaches 1 as n is increased ; 
the limiting value of G,(%,, $,) is then 


limit G(R, Pn) =(" et Beas 
0 Cos 


= (a, + B,) log tan fi + =) —B, singdn. [6] 


By Art. 180, [5] and [6], 


limit &,, Vee log tan 5 os 4 F'(k, ¢$). 


[4] can thus be written 


G (k, s=F Oh 8) a—F(1 +5 +24 


Qnr—2 gn-1 
ae ear ae 


+8 | sing +— singh + 





ae ih 








2 ging, +. 
eo OL sin nee 
ki, ic |) ieee alee 
Qnr-1 on 


+ —__——- sin¢,, _1 — ——_————_ sing, |. 7 
VER, «Biya © Mei... Baa + | His 


Cuap. XVL.j ELLIPTIC INTEGRALS. 2295 


Ifa=1, and B=—k’, [7] reduces to 


H(k, $) =F #)| 1+ (142 ce 


ue 
Qn—2 Qn-1 
Se a Pe) 








=A) sing + sin ds + ——. gin dy + «+s 
Vik iy 
+ ait sing zl sing ] [8] 
2) ae [ed] a Spo ars ene ee rE n 19 C 
AY Ky in Kn oo Vik, ...Ky 4 
where k, = tl , and sin (2¢,—4¢,-1)=, sing, [9] 
‘p—-1 


By Formulas [8] and [9] an Elliptic Integral of the Second 
Class may be computed without difficulty. 


184. Formula [2], Art. 183, may be used to decrease the 
modulus and increase the amplitude of an Elliptic Integral. 
Interchanging the subscripts, we have 





G (k, o)= : + fy | (Ay, di) — a sin #1) ) 








or, since e —, (Art. 183 [8]), 

E(k, ¢)=! sel (I, dn) — 8 5 sings b [1] 
where 
Bp age M8) =VIRF tang, aa t5, Bia te 


[2] 


226 INTEGRAL CALCULUS. (ART. 184. 


oo 


A repeated use of [1] gives 


' Lik ey ale odo tee 
G (k, ¢= 5 Ts. EAD Glas bn) 











=p ase £8 sind; + ital , ~ +" 8, sin d+ 





2 2 
1+h 1l+h 1i+k ; 
™ ( ss te eo leg af * 8, 18inde |, 13] 
2 2 2 
where oe ee 





and g=atdR(l +d 24 Aah 4 ae 


Just as in Art. 181 limit &%, =0, therefore limit 8, =0 and 
limit G, (Kn $n) = dd = an hy: 
eC 


By Art. 181, [2], ws ; 1th tthe, = FG, ). 




















By Art. 180, [5 | 
ad Dd tis po ao, 


Hence [3] becomes 


G0 d)=F Uh 8) e+ 4B(145 Bs Hale 5 Malad )] 


Re + Vi sin atu + YETI gin gy 4» } 4] 





sin d; + 


If a=1 and B=—#’, [4] reduces to 


Cuap. XVI.] ELLIPTIC INTEGRALS. 227 


E(k, $) =F (h, [1-50 +2, hike, Tatas +. )) 


Be sind; + — sin d, + —+=—= Vike ki 7 aks sind, + af [5] 


here 
Hen t—ie 


oy yt ne Tiaielneaalline sal 
ee 


Pia 


and tan (¢, — $1) = V1 —K,_, + tang, +. 


[6] 
We have seen in Art. 181 that if o=5 db, = 2 x. 


Therefore, for a complete Elliptic Integral of the second 
class we have 


aod aL Herb Peee yo 


Formulas [5] and [7] are admirably a to computation. 
We give in the next article the work of computing 


E V2 2 by each of the methods just given, and of com- 
2° 4 i 


puting u(y s) by the second method; using, as far as 
possible, the values already employed or obtained in Art. 182. 


185. a(S a} Metuop or Art. 183. 


Here, as we have seen in Art. 182, if we carry the work only 
to five decimal places, k,=1, and our working formula will be 


E(k, $) =F (k,¢) E +e(2 Be a 


1 
2 
2. sin +: 
kk, 





ay 
—k| sind +——sin gd, — 
[ VE 


228 


log 2 = 0.30103 
log k = 9.84949 
colog ky, = 0.00649 


0.15701 


log (1 diy fa a) = 9.43391 


1 


V2 2 


log F(*S = 9.91701 


9.35092 


logk = 9.84949 


log sind = 9.84949 


9.69898 


log 2 = 0.30103 
log Vk = 9.92474 


log sin ¢; = 9.78445 


0.01022 


log 2? = 0.60206 
log Vk = 9.92474 
colog Vk, = 0.00324 


log sin dy = 9.78122 


0.31126 


—k (sing + = sings — 


INTEGRAL CALCULUS. 


[ART. 185. 


2k 
1 


1+k=1.70712 


14+k—=" = 0.27159 


1 


= 1.48553 


(3, a +h- 2) = 0.22485 
2°4 Te, 


> 


k sing = 0.5 


2k sind, ieee 
Vie 


2 
PK sin by = 2.0477 


i 











<raitt $s) = 0.5239 
1 


/2 x 2k 
14+eo NS 0 eae 
Fle D( a = ete 


E (2 i= — 0.74825 


Cuap. XVI.] ELLIPTIC: INTEGRALS. 








229 
= ~ 
E (2 t) METHOD o* ARK. 184. 
k,=0. Therefore our formula is 
Ee i” ‘i k ss 
E(k, ¢) =F(k, *)!, fe e pe a 
es (i sin d, + als Uy sin d ») 
log k, = 9.23443 
log ky = 7.87246 
colog 4 = 9.39794 
| 6.50483 ae = 0.00032 
Kt — 0.08578 
2 


1+2 a4 A ks — 1.08610 


w/e wlS 


log 0.728475 = 9.862415 1— (3 +4 os) 


log F(“S = = 9.91700 
2° 4 


(tae eee ae — 0.271525 


= 0.728475 


9.779415 ae 5) (0.728475) = 0.60178 


log k = 9.84949 

log Vk, = 9.61722 
colog 2 = 9.69897 
log sin $; = 9.99370 


9.15938 4 


= 0.14434 


230 INTEGRAL CALCULUS. [ART. 185. 


logk = 9.84949 
Peale = = 9.6% (22 
log Vip == 8.93623 
colog'4 = = 9.39794 
log sin @, = 9.52592 SOs 
ERS 2a kV iy Ko _- 
7.382680 sha sin d, = 0.00212 


k ( sind; + on Vii Is sin tn) = 0.14646 


r(S " (0.728475) = 0.60178 


—_—<—__——— 


2(% i= 0.74824 


2(% =) Meruop or Art. 184. 


5 
E(k,7\=E(k,= 1-f 142 14 oe 
5 2 
2 A 
1 ~ ( Hecate 7 = 0.728475 log 0.728475 = 9.862415 


log (S 5)= 0.26813 


2(4 5)= 1.3507 log E (2 3)= 0.18054 





Cuap. XVI.] ELLIPTIC INTEGRALS. 231 


186. An Elliptic Integral of the first or second class, whose 
amplitude is greater than 5° can be made to depend upon one 
whose amplitude is less than a and upon the corresponding 


Complete Elliptic Integral. 


We have 


F(k, x)= f = ae inte ie K+ j by [4], Art. 179. 


In 


nla > 


"dd sat ale 
Ae let @ W; 


then dé =—dy and Ad= V1 —F’sin?'d = V1 — FB sin’y = Ay, 


and we have “db ak he dy _ is ee is 
rAd 
2 
Hence F(k, 7) = Aine eK: [1] 
ni+p 


F'(k, nw +p) = 


= (ies + fits fe wet ee + fi 
0 


(p+1)t 
Su lett Pb=pr+y; then dd= dy, and Ad= Ay, 


pr 





"add 


and we have ie fa - (i 2K. 


232 INTEGRAL CALCULUS. (ART. 186. 


n+p 


The substitution of w for ¢ — vz in f ve gives us 


n+p p 


p 
dp (a a aoe 
ike | AW is AB as 


Therefore F(k, nr +p)=2nk+F(k, p). [2] 


In like manner it can be proved that 


F (k, nx — p)=2nK — F'(k, p), [3] 
E(k, nr+ p)=2nH+ E (k, p)s [4] 
E (k, nz — p) =2nE — E(k, p), [5] 


where B= E(k 4 is the complete Elliptic Integral of the 
second class. ; 


A table giving the values of the Elliptic Integrals of the 
first and second classes for values of the amplitude between 


0 and 3 is, then, a complete table. 


Such a table, carried out to ten decimal places, is given by 
Legendre in his ‘* Traité des Fonctions Elliptiques.” We give 
in the next article a small three-place table. 

It must be noted that the first column gives F'(0, ¢) and 


E (0, #), that is, ‘fi see @; and that the last column gives 
0 
F(1, ¢) and E(1, ¢$), that is, log tan (F+$)and sin d. 


The complete Elliptic Integrals, 


ae P(K 2 ane H(t, 5) 


are given in the last line of each table. 


at: 
* 


INTEGRALS. 


ELLIPTIC 


Cuap. XVI.] 






























































F'(k, $). 

k=0° #£=01 k=02 k=03 | k=04 £=056 £=06 | k=07 k£=08 £=09 k=1. 

p sin 0° sin 6° sin 12° sin 18° | sin 24° sin 30° sin 87° | sin 45° sin 53° sin 64° sin 90° 
0° 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 
5 0.087 0.087 0.087 0.087 0.087 0.087 0.087 0.087 0.087 0.087 0.087 
10 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 
15° 0.262 0.262 0.262 0.262 0.262 0.263 0.263 0.263 0.264 0.264 0.265 
20 0.349 0.349 0.349 0.350 0.350 0.351 0.352 0.353 0.354 0.355 0.356 
25 0.436 0.436 0.437 0.438 0.439 0.440 0.441 0.443 0.445 0.448 0.451 
30° 0.524 0.524 0.525 0.526 0.527 0.529 0.532 0.536 0.539 0.544 0.549 
35 0.611 0.611 0.612 0.614 0.617 0.620 0.624 0.630 0.636 0.644 0.653 
40 0.698 0.699 0.700 0.703 0.707 OFZ 0.718 0.727 0.736 0.748 0.763 
45° 0.785 0.786 0.789 0.792 0.798 0.804 0.814 0.826 0.839 0.858 0.881 
50 0.873 0.874 0.877 0.882 0.889 0.898 0.911 0.928 0.947 0.974 1.011 
55 0.960 0.961 0.965 0.972 0.981 0.993 1.010 1.034 1.060 1.099 lis4 
60° 1.047 1.049 1.054 1.062 1.074 1,090— "ded 2 1.142 1.178 1.233 Laké, 
65 1.134 AY 1.143 1.153 1.168 1.187 1.215 1.254 1.302 1.377 1.506 
70 tone 1.224 1.232 1.244 1.262 1.285 1.320 1.370 1.431 1.534 1.735 
rie 1.309 1312 1.321 1.336 ho 57 1.385 1.426 1.488 1.566 1.703 2.028 
80 1.396 1.400 1.410 TAz7 1.452 1.485 1.534 1.608 1.705 1.885 2.436 
85 1.484 1.487 1A99 1:5 19) 1,547 1.585 1.643 LJal 1.848 2.077 G4: 

K290° 1.571 1.575 1.588 1.610 1.643 1.686 1752 1.854 1.993 2.275 oe) 


| 
See NS eR ed SO a oa ie Se ee OE ase Re pee eee Se SA NS ee Me ee i ok Se eS ee 





[ ART. 187. 


INTEGRAL CALCULUS. 

































































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Cuap. XVI] ELLIPTIC INTEGRALS. 235 


Addition Formulas. 


188. The Elliptic Integrals, F'(k, x) and E(k, x), may be 
regarded as new functions of x, defined by the aid of definite 
integrals ; namely, 


x dx 
ye 4), Vil — 2) (1 — Kae) 


x — i? 2 
E(k, 2)={",| oe - ax; 


see Art. 178, [1] and [2]. 

We have seen how we may compute their values to any 
required degree of approximation when & and @ are given. 
It remains to study their properties. 

We are familiar with other and much simpler functions which 
may be defined as definite integrals, and whose most important 
properties can be deduced from these definitions. 








For example, we may define log as afi Mes sin-lxw as 
By 
x dx 7 ae dx 
, tan “x as 

oV1—2 o1+2 
tions may be based upon these definitions. For instance, the 
fundamental property of the logarithm is expressed by what is 
called the addition formula, 





, and the theory of these func- 








loga« + logy = log (ay), 


and the whole theory of logarithms may be based on this 
property ; and there are addition formulas for the other func- 
tions defined above; namely, 


sin-w + sin y = sin? (aV1— y+ yvV1 — 2), 





tanta +tan7y= tan-*( Este ) 
1 — xy 


236 INTEGRAL CALCULUS. [ART. 188. 


These three important formulas are usually obtained by more 
or less elaborate methods involving the theory of the functions 
which are the inverse or anti-functions of the log, the sin-'a, 
and the tan-‘a, that is, of e”, sina, and tanaw; but they may 
be obtained without difficulty from the definitions of log 2, 
sin-tz, and tan~'w, as definite integrals. 


Take first loge = f =. 
Lea 
Let us determine y in terms of x, so that 
logx + logy = loge, (1) 
where c is a given constant. 
Since logy ={°%, 
Noe 


if we differentiate (1), we have 


de, dy _o, 
a: Natt 
or ydz + ady = 0. (2) 


Integrate (2), and we get 
fyae + J ady=C. (3) 
Simplify the first member of (3) by integration by parts ; 
ay — { ady + ay — { yde= C; 
or 2 ny — { (2dy + ydx) = C. 


Reducing by the aid of (2),  2ay=C, 


or ay = Ch, (4) 


Cuap. XVI.] ELLIPTIC INTEGRALS. 937 
where C, is an undetermined constant. To determine (,, let 
% = 1 in (4), and we have y = C,; when «=1; let e=1 in (1), 
ld» 
then loge = f ae 0, logy=loge, and y=c, when x=1. 
1: 2 | 
Therefore, C}=c, and wzy=c. Consequently y= “ ig the 
2 
required value of y, and we have (1) 
logxw + log® = loge. 
0 

, We can express this relation more neatly by replacing c by 
its value zy, and thus we reach our required addition formula 


logx + log y = log (ay). [5] 


189. The addition formula for the sin can be deduced in 
exactly the same way. We wish to determine y so that 




















sin-'w-+siny = sin“te. (1) 
We have sin te= f° holt , sin ly ={. dy 
° V1 — 0 1 — y? 
Differentiate (1). 
clac dy 
ete aid (2) 
Vice Al 
or Vi—y- dx+ V1 —a? - dy=0, 


fyi —y’. d+ {V1 —2-.-dy=C. 
Integrate by parts, and 


eVl—y +y V1 — 2? + eee te 5) =O: 
= —y 











or, reducing by (2), 


aVi—y+yV1—#=C. (3) 


238 : INTEGRAL CALCULUS. (ArT. 189. 


To determine C, we have from (8) y=C when x=0, and 





é =(, when 





0 

from (1) y=c when a= 0, since sin"'a = = 
0 9/1 

a=0. Hence C=c, and «V1 —7?+y~V1— a? =c, and, finally, 


sin-!e + sin-y = sin“! (aV1—y? + y V1 — 2”). [4] 


To get an addition formula for the tan™, a slight device is 
required, that of dividing the differential equation correspond- 
ing to (2) by 1— a7’. 

As before, let 











tan-“'x + tan“'y = tan !c, (5) 
where farsa a= [pee 9 
afO 2 oe H bs 
and tan y= {~ be. 
o1l+y 
dx dy 
Chad s =0 
14+2 = Lia 
or (1+ 7") dw+(1+27) dy=0. (6) 


Divide by 1 —x’7’ and integrate. 
2tY dwt fAt*,. dy=O 
Upper et 1 — xy/ Yaa 


Integrate by parts. We have 











; Ly oe 2y 1 > 2 
oS alae Ce can +2) dy + yee 
tae ash peas! da + xy (1 +27) dy], 


1— 27 mcr oS va 
and 

1+/7 1+ _ oy 
1—2vy 1—ay? 1—ay 








Cnap. XVI.] ELLIPTIC INTEGRALS. 239 


Hence 





PET Ys 2ay (1+ ay) 1 2 ) a 
ey ay? [i +7)da+ (1 +2’) dyj= 0. 





Therefore, by (6 aS OE iC. 7 
erefore, by (6), penn C CP) 


To determine C, we have from (7) y=C when «=0, and 





0 
from (5) y=c when x=0, since tan “x = ; ee ;=0 when 
ol—@ 
oO, 
Hence C=c, and 2+Y alg, 
1 ~— avy 
and, finally, tan-'a + tan”-'y = tan” aa [8] 


190. To get.an addition formula for &(k, x), as before 














let F (k, ©) +F (k, y) = F(k, ¢), (1) 
: lx 
here I (by) = : : 
2 Loe f} / (1 — a?) (1 — a?) 
y dy 
J Va-y)a-Fy) 
os ab ee res (2) 
V—#)(1—#a?) V(1—y)(1— Fy’) 
or 


Vi — 9)(1 — ky") - da+ V1 — a?) (1 — #2?) -dy=0. (8) 
Divide by 1 — k’2’y’ and integrate. 


ee 2) (kia Ch sheet yi 
1) gchar — hey’ ace otf 1— kay? eee 


240 INTEGRAL CALCULUS. (ArT. 190. 


Integrate by parts. We have 





V (1 —y?) (1 — ky?) y 
i ley | att ee 








b ° dy 
—(1+%?)(1 +h 2? y’) | 4 
Vi) (1 — 4) 


+ 2h ayV (1 — y) (1 — ky") - da? ; 





q VG=“)0— Pe) _ a 


TAL ert Tee Jeet Dee 2 ie an? 2 
1 Baty Gakayy ae 


2 dx 
= (1 + k?) ¢! + kx? y?) )§ 
V (1 — 2%) (1 =a) 
+ 2h ay V (1 — 2?) (1 — K2°) - dy?- 


Hence 


aV (1—7*) 1 + yV A — &) 0 — 2") 
1 — ha? 7? 








cllneaeaae [2k?(a?+ y?) — (1 +k?) (1 + Way’) | 


a 
N(— 2?) (1— iat) VT 98) (1 a 





+ 2hay[V(1—y*) (1 — Ry’) - da 





+ V(1—2*)(1— Fa"). dyt = C. 
Reducing, by the aid of (2) and (3), we have 


eV (1—y*) 1 y) +yV (1 — 2?) d — Ba?) _ 
1—Roy? =C. (4) 


Crap. XVI.] ELLIPTIC INTEGRALS. 241 


To determine CO, from (4) y=C when «=0, and from (1) 
y=cwhenv=0. Therefore (=c, and we get 


PF (k, x) 4- F(R, y) 


" (ik, is CESS a Solas), £5] 
7 ¥ 


our required addition formula. 

An addition formula for E(k, x) can be obtained in very 
much the same way, but the work is rather complicated, and 
it is better to use a method which will be explained later. 


THE ELLIPTIC FUNCTIONS. 


191. We have just seen that there is an analogy between 
the Elliptic Integral Fk, x), and the familiar functions log2, 
sin +a, and tan-'a; and we know that the theory of these 
functions is ultimately connected with that of their inverse 
functions, log~'w or e", sinw, and tanu; and, indeed, that the 
latter are so much simpler than the former that it is customary 
to regard them as the direct functions, and the logarithm, the 
anti-sine, and the anti-tangent as the inverse functions. 

For example: the first three addition formulas just obtained 
are much simpler when we express them in terms of the direct 
functions, and they become 


log (vu +v)=logu- logy, 


or ahs =e". @, [1] 
sin(u+v) =sinuV1—sin?v + sinv V1 — sin’; 

or sin(w+v) =sinuwcosv+cosusiny, [2] 
tan(w+v) = tanwu-+tanv, [3] 


1—tanuw-tanv’ 


and in this form they seem to better deserve the name of 
addition formulas. 


242 INTEGRAL CALCULUS. (ArT. 192. 


In the same way the addition formula for F'(k, %) can be 
more simply written in terms of the function which we might 
naturally represent by F ‘wu (mod. k); and, as we might 
expect, this function has many interesting and important prop- 
erties which well deserve investigation. 

Since in most of the work which follows we shall generally 
employ the same modulus throughout, we shall not take the 
trouble to write it except in the few cases where its omission 
might give rise to confusion, and then we shall put (mod. k) 
after the function, as above with F’~'u (mod. k). 


192. In Arts. 178 and 179 we have adopted two forms of 
notation for an Elliptic Integral of iy first class, & (k, x) and 


P(k, $) 3 


F(k, «)= <i lame ea 
has J, Ni 


odd. 
POs $)= f Ga eas =ih Ag’ 








where z= sind, V1—2°=cos¢, 
and V1 — Ret = V1 — Fk? sin?’ d@ = Ad. 
If we let ah es Bie as) ae ice Ne 


we have in Art. 179 called @ the amplitude of uw, and sing, 
cos ¢, and Ad may be called the sine, the cosine, and the delta of 
the amplitude of w; and ¢, sing, cos¢, and Ad may be written 
amu, sinamw, cosamw, and Aamzw, or, more briefly, amu, sn, 
cnu, and dnu; and may be read amplitude wu, sine amplitude uw, 
cosine amplitudeuw, and delta amplitudew. Formulating, we 


have 
“a= F(k,0) =P (ho), | 


d= amu, 
Co Bp she rit [1] 
V1 —a = cosd = cnu, 
V1 —e’? =Ad=adnu, 


Cuap. XVI.] ELLIPTIC INTEGRALS, 243 


snu, cnu, dnwu, are trigonometric functions of ¢, the ampli- 
tude of wu, but they may be regarded as new and somewhat 
complicated functions of wu itself, and from this point of view 
they are called Elliptic Functions of u. 

amw also is sometimes called an Elliptic Function ; and there 
are various allied functions that are sometimes included under 
the general title of Elliptic Functions. We shall, however, 
restrict the name to snu,cnu,and dnu. They have an analogy 
with trigonometric functions, and have a theory which closely 
resembles that, of trigonometric functions, and which we shall 
‘proceed to develop. It must, however, be kept in mind that 
the independent variable wu is not an angle, as in the case 
of the trigonometric functions. 

Of course, with our notation, u= F'(k, x) =sn-'x (mod k),. 
or u=F(k, ¢) =am'¢ (mod k). 

The fundamental formulas connecting the Elliptic Functions: 
of a single quantity follow immediately from the definitions: 
[1], and are 











sn?u+en?u=1, (2), 
dn’u+ k’sn?u=1, [3] 
dam wu 
= dnu 4 
du ; Ls 
dsnu 
=cnu.dnu 5) 
du ‘ / : Ba 
ee snu.dnu, [6] 
du 
ddnwu 2 
=—k*snu.cnu, [7] 





du 


The only one of this set which needs any explanation is [4]. 


dd 
h — i 5 
We have U= ; —— 


244 


hence 


and, finally, 


Since 


we see that 


INTEGRAL CALCULUS. 


du = 


damu 
du 


Ae vil 


Gps du = 
Ag’ 


= dnu 


bd(—$) _ 


A(— 4) 


damv , 


dnu 


ey 


am(— wv)=— amu, | 


sn(—u) =—snu, 
cn (—wu) =cnu, 
dn (— wu) =dnu, 


a 
| 


ope) 


td 


Ad’ 


[ART. 193. 


[3] 


That is, snw is an odd function of ~, and cnu and dnw are 
even functions of wu. 


Since 


we have 


figeo 
Dues NN 
Bes 
sn{0)-= 0, 


dn((t) jae is 


[92 


193. Our addition formula for the sine amplitude flows 
Let u=F(k, 2) and 
v=F(k,y), and take the sine amplitude of each member of 
[5], Art. 190; we get 


immediately from [5], Art. 


sn(u+v) 


0 aay 


190. 


__snu.cny.dnv+cnu.snv.dnu 
sn?u.sn?v 


If now we replace v by —v, and simplify by [8], Art. 192, 


we have 


sn(w—v) = 


snwu.cn 


1—k’. 


y.dny — cnw.snv. 
sn?u.sn?v 


and the two formulas can be combined if we use the sign + ; 


Cuap. XVI.]. ELLIPTIC, INTEGRALS. 245 


snwu.cnuv.dnvtcnu.snv.dnu 

sn (ut v) = ——— 1 
( ) 1—k’?. sn?u. sn?v [1] 
From [1], with the aid of [2] and [3], Art. 191, we can get, 

after a rather elaborate reduction, the addition formulas for 

en and dn. | 





cnu.cnv + snu.snv.dnw.dnv 
CLE. C) = ; 9 
( ) Tok ent . any [2] 
Ae 
dn (u + v) _dnu.dnv Fk -snu.snv.cnu.cnv [3] 





1 — k?. sn?u.sn?v 


From [1], [2], and [3] a large number of formulas can be 
readily obtained. We give only those for sn; there are 
similar ones for cn and dn. 


_ 2snu.cnv.dnv [4] 
1 —k?. sn’?u. sn*v 





sn(u+v)+sn(u—v) 


2enu.snv.dnu 
sn (w+v)—sn(u—v fm pee biatene hash Renter eam 5 
oat) ( ) 1 — k?. sn?u. sn7v (2 
sn (u+v).sn(u—v) pores Stu 80 Ub [6] 
1 —k?sn?u. sn?v 

an? 2u.dn?v 
1+sn(u+v).sn(u—v TNs Mea EEC 7 
ie We +2) ( ) 1—k’.sn?u.sn?v u¢] 


, dn?v + k?. sn?u. en?v 
Peer en (ut ) sn (e—v) = ————________—___.. 8 
ik ae ( ) 1—k’*.sn?w.sn?v [8] 


(cnv +snu.dnv)* [9] 
1— k?. sn?u.sn?v 


[1+sn(u+v)][1+sn(u—v)]= 


From [2] and [3] come the useful formula 


cn(wu+v)=cnu.cnv—snuw.snv.dn(wu+v). [10] 


246 INTEGRAL CALCULUS. (ART. 194. 


194. If in formulas [1], [2], and [3] of Art. 193 we let 
v=u, we get the following formulas for sn2u, cn2u, and 
dn2u: f 

2snu.cenu.dnu 








on Ji = eee 1 
1—sntu  ’ Be 
cn?u—sn?u.dn?u 1—2sn?u+k’sntu 

cn2u = re ee Be eo ee 2 

1 — ena 1 — ain ‘ [?] 
dn?u—k?.sn*u.cn?u 1—2k'sn?u +k’ sn*u 

a Te W LY Recor yp sore enblaotsxer evos elaaibecece Shy SS ; 3 

Le anita 1—k’sn*tu Bee 


From these come readily 


9sn?u.dn?u 
1 cn = -- 4 
1—k?shtu ’ bal 
1+en2u = Mess ah [5] 
t—Jesn*ta 
- F, 2. 2 
1—dn2y— 2% su. cui [6] 
1—k’sn*u 
2 dn? 
1+ dn Ju aia 
BY 1 —<' sn*u a 


195. Replacing u by and dividing [4] by [7] and [6] 
by [5], Art. 194, we have 








Ub a SC hy ef —dnwu [1] 
201 edna (La tn) 
snag g 2 
on?! = dnu + chu _ k Ben ecnu +dnu [2] 
2 1+ dnu k(1 + cnw) 
Ae] 2 2 
an?’ — +dnu+k cnu __k RSE dnw) [3] 
2 1+ dnw k*(1 + enw) 


where k'?=1—k’, and is the square of the complementary 
modulus. 


Cuap. XVI.] ELLIPTIC INTEGRALS. 247 


From [1], [2], and [3], we can get without difficulty the set 


gt dnu —cnu 
(es oma ‘ ’ 
2 k'?+dnu—k’?cnu 





sn 


[4] 


ot —s—sédK#(1 + enw) 
2 k?+dnu—kenw 





cn 


[5] 


ley at ki*(1+dnu) [6] 
2 k?+dnu—Kenu 





Numerous additional formulas can be obtained by the exer- 
cise of a little ingenuity, but we have given the most useful and 
important ones, and they form a set as complete as the usual 
collections of trigonometric formulas. 


Periodicity of the Elliptic Functions. 
196. We have seen (Art. 186, [2]) that 
F(k, nx +p)=2nkKk+F (kh, p), [1] 


where A is the complete Elliptic Integral of the first class. 
Let u= F'(k, p), and take the amplitude of each member of 
[1]; we get 
am (u+2nK)=n7r+amu ; [2] 


or, replacing n by 2n, 


am (u+4nH)=2n7r+ amu; [3 | 


whence 


sn(u+4nAK) =snu, ) 
en(u+t4nk) =cnu, t : [4] 


dn(u+4nK)=dny, J 
and snu, env, dnw have periodic functions, and have the real 


period 4A. dnw actually has the smaller period 2K, as may 
be seen by taking the delta of both members of [2]. 


248 INTEGRAL CALCULUS. (ArT. 196. 


Since the amplitude of X is 3 we have 


snifties 1 
cnK =0, \ [5] 
dn kK ='k'. J 


and our addition formulas [1], [2], [8], Art. 193, give us 
readily 








sn(u+tA) = ae | 
k'snu 
Kk =o ~s 
cn(u+ I) ane i> [6] 
k! 
dn(u+h) = a 
nu 


sn(u+2K)=—snu, 
cn(u+2A)=—cnu, |, [7] 
dn(u+2k)= dnu, j 








sn(u+3K)=— 2%, 
dnu 
cn(u+3kh)= kisnw [s 
dnu ! : 
k! 
dn(u+3k)= ; 
dnu 


sn(u+t4k)=snu, 
en(u+4H)=cnu, $, [9] 
dn(u+4k)=dnu, i 


a confirmation of [4]. 


Cuap. XVI.] ELLIPTIC INTEGRALS. 249 


197. It is easy to get formulas for the sn, cn, and dn of an 
imaginary variable, wV — 1, by the aid of a transformation due 
to Jacobi. 


Let v= F(k, =f boca (1) 


so that 6=amv, sind=snv, and cosd¢=cnv. In (1), re- 
place @ by ¥, @ and yw being connected by the relation 


sind = V—1. tany, (2) 
whence COS d= secy, (3) 
ee hain dé = 1 Re tante, (4) 

and db = V —1.secw. dy. 


Since w and ¢ equal zero together, 


v= (2 ib Bey ¥ secydy _ 
V1+ Kk tan?y 





Rae tc (ity EO Nilo ar Gilet.) 
oe ue 


If now we let u= F'(k', Ww), 
we have v=uv—i. (5) 
Hence, since Y=amu(modk’'), we have from (2), (3), 
and (4), 
Soe | 
J snu crude) 
cnu (mod k’) 
eae ar 
cnu (modk')’ 


snv (modk) = 


env (modk) = 


dnu (mod hk’) , 


AE a enu (mod k’) ’ 


or, asu=uvV—1, 


250 INTEGRAL CALCULUS. (Arr. 197. 


14s sea 25, k') 
Vd cp —4/ Shu (mod 
main mene? enu (mod k')’ 


; Me [6] 


cnuy —l (nod Hie eee, 
i \ ) enw (mod k’) 


i +f 
dnwV—1(modk) = dnw (modki ) 
enw (mod k') 





It is interesting to note that if uw is replaced in (6) by 
u\/ —1, the formulas reduce to 
sn(— wv) =—snu, 
en(—w) =cnu, 
dn (—u)=dnu, 


and are still true. Consequently, in (6), uw may be either a 


real or a pure imaginary. 


Tv 


S| 4 


2 > 
Let {—a- eA — Agi 
0 Ay (modk') 0/1 —k? snp 


Then, by Art. 196, 44’ is a period for snw(modk’), 
enw (modk'), and dnu(modhk’). 


Hence Naa Tee 
sn (uv — i+4nK'V— (i R snuvV— 1, 


on (uV—1+4nK'V—1) =cenuvV—1, 
dn (u VJ—1+4nK! ee =dnuV—1; 
or, replacing Fa eels hi by v, 
sn(v+4nK'V—1) =sny, | 
cen(v+4nK'V—1)=cnv, }, [7] 
dn(vt4nk'V—1)= dn, | 


and 4.A4'-/ —1 is a period for sn, cn, and dn. 

We see, then, that our Elliptic Functions, like Trigonometric 
Functions, have a real period, and, like Logarithmic Functions, 
have a pure imaginary period. They are, then, what may be called 


Cuapr. XVI.] ELLIPTIC INTEGRALS. 251 


Doubly Periodic Functions, and they are often studied from the 
point of view of their double periodicity. 

Like Trigonometric Functions, the Elliptic Functions may be 
developed in series, and from these series their values may be 
computed, and tables resembling Trigonometric tables may be 
prepared. | 

A partial three-place table is here presented as a sample. It 
is complete for Elliptic Functions having the modulus Me ; that 
ide Ost. 


Moputvs ae ser 


u sn wu en u dn u 
0.00 0.000 1.000 1.000 
0.05 0.051 0.999 0.999 
0.15 0.150 0.989 0.994 
0.25 0.247 0.969 0.985 
0.35 _ 0.340 0.940 0.971 
0.45 0.429 0.903 0.953 
0.55 0.512 0.859 0.932 
0.65 0.589 0.808 0.909 
0.75 0.659 0.752 0.885 
0.85 0.722 0.692 0.860 
0.95 0.778 0.628 0.835 
1.05 0.827 0.562 0.811 
1.15 | 0.869 0.494 0.789 
1.25 | 0.906 0.424 0.768 
1.35 0.935 0.353 0.750 
1.45 0.959 0.284 0.735 
1.55 | 0.977 0.213 0.723 
1.65 | 0.990 0.143 0.714. 
1.75 0.997 0.072 0.709 

K.1.85 1.000 0.000 0.707 


From this table, by the aid of formulas [4], [6], [7], and 
(8) of Art. 196, snu, enw, and dnuw may be readily obtained 
V2 


for any value of u if the modulus is Patt 


252 INTEGRAL CALCULUS. [ART. 198. 


As a matter of fact no complete set of tables for the Elliptic 
Functions has been published, and their values are usually ob- 
tained indirectly from Legendre’s Tables of Elliptic Integrals 
(v. Arts. 186, 187), unless especial accuracy is required, in 
which case they must be computed by methods which we have 
not space to give. 


198. The Elliptic Integral of the second class # (kh, $) can 
be expressed in terms of Elliptic Functions, and for some 
purposes there is a decided advantage in the new form. 


wh 
We have E(k, ¢) = { Ad.dd. 


Let u= F'(k, $), then ¢ = amu, and, E (k, d) may be written 
E(k, amu), or, more simply, H(amw), if the modulus can be 
omitted without danger of confusion. 


Then E (amu) = (oda wu.damu; 
0 
or, since by (4), Art. 190, 
damu=dnu.du, 


E (am) = {an‘u du. [1] 


As an example of the usefulness of the form just given in 
[1], we will employ it in getting an addition formula for 
Elliptic Integrals of the second class. 


E (amu) + # (amv) 


= { dntw . du + {ant .dv 

0 0 

a 4) "an?z . dz +- f “dn2z . dz 
; 0 0 


u+v v u+v 
=f dn?z.dz + 4 dn’?z.dz— f dn’z. dz 
0 0 e/t 


= H[am(u+v) | + dn*z . dz — { dn'z . dz. 
0 u 


Cuap. XVI.] ELLIPTIC INTEGRALS. 253 


Replacing z by w+ 2, and remembering that wu and v are given 


constants, 
ut+v v 
if dn?z. dz ={ dn? (u +2) dz, 
uU 0 


and 
E (amv) + # (amv) = 
E{am(u+v)|— (Tan? (u+z) —dn?z] dz. (2) 


dn? (wu +2)—dn?z= [dn (wu +2)+dnz][dn(u+z2)—dnz]. (3) 


We can obtain from [3], Art. 193, the following formulas 
analogous to [4] and [5], Art. 193, 


2dnu.dnv (4) 
9 
— k*sn?u.sn?v 





Gata?) + dn(u —v)= 5 





2k’ snu.snv.cnu.cnv (5) 
1 —k*sn?u. sn?v 


dn(w+v)—dn(u—v)=— 
If in (4) and (5) we let u+vu=u+2, and u—v=z, and 
substitute the results in (3), we get 
dn? (u +2) — dn’z 


Sant i he von ane 
anton (5 +2)en(5+2)an(5 +2)sn5 en5 any 


i 1 —22sn? sn2/ & j 
| on'5ou'(S +2) 


and 


fan? (u +2) —dn’z]| dz 


zien (3+ en(5 +2) an(5 +2)ae 
ok ee 2 2 2 
= — 2sn—cn—dn= § —— Ss JH 
202 2 1 — Kan? sn? 5+) 
2 2 
9sn on an” 
2 2 2 1 
sn? 1—Ren2ign?(Y +2 
2 2 2 


254 INTEGRAL CALCULUS. [ART. 199. 
since — 2k? su sn & + “| cn ee :) dn € + “) dz is the differ- 


ential of 1—k? sn’ sn’ ¢ + ‘) 


f, [an (u +2) — dn’z] dz 


LEAN IGL AL ee Rt o/u u 
fe 2ehe ons an an'(' an ») — gn? 


1—ksnt™ 1 — k? sn? sn? ay v 
2 2 2 | 
=—k’.snu.snv.sn(u+v), 


by (1), Art. 194, and [6], Art. 193. 
Hence by (2), 


E(amu)+ £ (amv) = H[{am(u+v)]+A’snu.snv.sn(u+ v), 
[6] 


our required addition formula. 
APPLICATIONS. 


Rectification of the Lemniscate. 


199. From the polar equation of the Lemniscate, 7?=a? cos 26, 
referred to its centre as origin and its axis as axis, we get as 
the length of the arc, measured from the vertex to any point, 
P, whose coordinates are 7 and 6. 


s=a eS Aa ee [1] 
0 /cos 26 0 JT—2 sind 


Cuap. XVI.] ELLIPTIC INTEGRALS. 295 


and for the are of the quadrant of the Lemniscate, that is, the 
arc from vertex to centre, 


T 


4 
a) (2] 


0 1/1 —2sin?6 


These differ from Elliptic Integrals of the first class only in 
that the coefficient of sin’@ is greater than unity, and they may 
be reduced to the standard form by a simple device. 

Introduce in [1] ¢ in place of 0, ¢ and @ being connected by 
the relation sin?¢ = 2 sin?d. 











Then we have sf] ~ sin’ é = cos d, v 
and dé = v2 _ cospdd 
2 /1—dsin’d 
Hence s= a as ¥ V2 mae $) [3] 
V1i—4sin’¢ d sin? 2 
and ey asf v2 (2 *), [4] 
V1—4 sin’? He 2 22 


The auxiliary angle ¢ is very easily constructed when the 
point P of the Lemniscate is given. We have r=avcos 20, 
and we have seen that Vcos 26 =cos@; hence r=acos¢. If, 
then, on a as a diameter we describe 
a semi-circumference, and with the 
centre O of the Lemniscate as a 
centre, and with a radius equal tor, 
we describe an are, and join with O 
the point @ where this arc intersects 
the semi-circumference, the angle made by OQ with a is equal 
tod. For 0Q=acos A0Q and OP= aVcos 26. 











256 INTEGRAL CALCULUS. (ART. 199. 


EXAMPLES. 


(1) Find the numerical value of a) SE 
0 V1—4sin’?¢ 
Ans. 0.843. 
6 dd noe 
(2) Reduce afi —_——*——. to an Elliptic Integral of the 
0 V1—nsin?d 
first class, when n > 1. 
¥ d 
Ans vy) nhs UAL 


| Tass hh 2 sin? 
Nn 


where sin’ y= sin’. 





(3) The half-axis of a Lemniscate is 2. What is the length 
of the are of a quadrant? of the arc from the vertex to the 
point whose polar angle is 30°? Ans. 2.622; 1.168. 


In the inverse problem of cutting off an arc of given 
length the Elliptic Functions are of service. As an interesting 
example, let us find the point which bisects the quadrantal are 
of the Lemniscate. 





Here ian av? 4 (2 ay 
2 2 eo 
and we wish to find ¢ and then 6. 
Let iret v2 7\; we need am~- 
2 iv2 2 


amu =>, snu=1, cnu=0, and dau =. 


By [1] and [2], Art. 195, 





nt = 1—cnu ou _dnu+cnu 
2 1+dnw’ 2 1+dnu 
Therefore, 
sn? ~ 
2 U 1— cnu 1 Ne 
= t= = ———— so = = 
on? & 2 dnu+cnu /2 V2. 


2 ae 


Cuap. XVI.] ELLIPTIC INTEGRALS. 257 
If, then, the required amplitude is 4, | 
tan’¢ = V2, 
and tang = V2. 


Since sin’?¢ = 2sin’?9, we can compute 6 without difficulty, 
and so get our required point. If, however, a construction will 
- suffice, a very simple one gives the point. 

Erect at A a perpendicular 
whose length is a mean pro- 
portional between a and avV2. . 
The angle subtended at O by 4 


this perpendicular is ¢, and ee 
the corresponding point, P, is Aaa 
found by the method described 9 Rat 
on page 255. 


. Rectification of the Ellipse. 


200. We have seen in Art. 177 that the length of an arc 
of an Ellipse measured from the end of the minor axis is 


piel 
s= | Noe - dx. [1] 
If we let x=asing, [1] becomes 


s=af°Vi—ésin’ $ .dp=aE(e,¢),. [2] 


e, the modulus of the Elliptic Integral, being the eccentricity 
of the Ellipse. If r=a, d= 53 and the length of the Elliptic 
quadrant is 





=a VI = @sing « a6 = a8 (e, 5); [3] 


258 INTEGRAL CALCULUS. [Arr. 201. 


The length of an arc of the Elliptic quadrant, not measured 
from the extremity of the minor axis, can of course be ex- 
pressed as the difference between two Elliptic Integrals of the 
second class. 

The amplitude ¢, corresponding to a given point P, of the 
Ellipse, is easily constructed as follows: On the major axis 

as diameter describe a circumference ; 

Ont eect extend the ordinate of P until it meets 

asing the circumference, and join the point of 
intersection with the centre of the ellipse. 
The angle the joining line makes with 
the minor axis is seen to be the required 
amplitude ¢. If ¢ is given, P may be 
6 = aa found by reversing the order of the steps 

of the construction. 





EXAMPLES. 
a 2 
The equation of an ellipse is i + e = 1, required the length 


of the quadrantal arc; of the arc whose extremities have the 
abscissas 2 and 2/2. Ans. 5.43; 0.944. 


(2) Find the abscissa of the end of the unit are measured 
2 2 
from the extremity of the minor axis in the ellipse oF a ks 
of the point which bisects the are of the quadrant. 
Ans. 0.9965 25 te 


201. By the aid of the addition formula 


E(amu)+ H(amv)= # [am (u+v)]+e?snusnvsn(u+v) 
({6], Art. 197) 


it is always possible to find an arc of an ellipse differing from 
the sum of two given arcs by. an expression which is algebraic 
in terms of the abscissas of the extremities of the three ares. 
This will be clearer if we modify slightly the form of our addi- 
tion formula. | 


Cnap. XVI.] ELLIPTIC INTEGRALS. 259 


Let g=amu, w=amv, andoc=am(u+v). 
Then the formula given above becomes 
E(k, 6)+ E(k, y)= E(k, o)+k’sindsingsine, [1] 
where ¢, w, and o are three angles connected by the relation 


coso = cos ¢ cosy — sing sinwAo, [2] 
by [10], Art. 193. 


If we multiply [1] by a and take & equal to e, we get 
aH (e,¢)+ak (e,~)=ak (e,c)+ on » Ly. Ley 


if %,, %, and x, are the abscissas of the points whose amplitudes 
are d, w, and o. 


. . * T . ° 
The most interesting case is when Ge aa which case: 


a 


aH(e, 7) is the arc of a quadrant. [2] then reduces to 


0 = cos ¢ cosy —sing siny V1—e’, 


lc ee ; 
or ” sin d sinw = cos¢ cosy, 
a 


or tan ¢d tany = a [3] 


and we get from [1] 
ak (e, >) -| aE («, ae ak (e, | =aesingsiny. [4] 


If, then, any point, P, is given, [3] will enable us to get 
the amplitude of a second point, Q, and - 
thus to find @, Q@ and P being so re- vs 
lated that the are BP, minus the are AQ, 
shall be equal to a quantity which is 
proportional to the product of the ab- 
scissas of P and Q. 0 <4 


260 INTEGRAL CALCULUS. [ART. 202. 


For the special case where ¢ and w are equal we have from 


[3], tand = [2, 
No 
and from [4], 
A are? 
BP — AP= ae’ sin’ ¢ = =a—b. 

a+b 
This point, which divides the quadrant into two arcs whose 
difference is equal to the difference between the semi-axes, has 


a number of curious properties, and is known as Fagnani’s 
point. 





EXAMPLES. 


(1) Show that the distance of the normal at Fagnani’s point, 
from the centre of the ellipse, is equal to a — b. 


(2) Show that the angle between the normals at P and Q in 
the figure is equal to Y—q@; that the normals are equidistant 
from O; that this distance is BP — AQ. 


Rectification of the Hyperbola. 


202. If the arc of the Hyperbola is measured from the 
vertex to any given point, P, whose codrdinates are « and y, 
its length is easily found to be 


2 2 
Ae te yy\ 
y 
(7) 11) 
0 y 
\ tag 
OE) sae 
y oe bt y 
or $= 7 dy, [2] 
I+ 


if e is the eccentricity of the Hyperbola. Let 


ae 
ary = tan d, 


Cuap. XVI.] ELLIPTIC INTEGRALS. 261 


= peo sec? ee dd 
V1 — Joints 


= sin® es 


and [2] becomes 


hence 


2  se’ddd == (esecede [3] 














0 Vi—Ksin?d “ve Ad 
if Ree 
e 
2 PY gh ay a2 2 
Re ee Lain’ — M cos’ 
A@ 1—-kh Ad 1-F Ad 
Z k? cos’ 
ral Meee | 
nd = —— « 2 A 1 i ( ae 
e "ae aS ace AOU a 
2 
pear [, sec? Ad dg — IF (k, 4) . 
Ge lL. — elas 


If we integrate by parts, 


eee bax CYA Sl eee 
il, sec bAddp=tangdad+h f Ad dd ; 


but pene is — Ad, 
he ee dp = F(k, $) — E(k, $). 
Hence 
“a a b? me 
= Ae i >) aise ay Ge [EH (k, 6) —tan¢ Ad]. 


But 1 — = ak 


262 INTEGRAL CALCULUS. (ART. 202. 


therefore he oF (k, o>) —aeH (k, ¢) + aetand Ad, 
ae 


or Se ie $) —aeEH ( 6) +aetand Ad. [4] 


The angle ¢ corresponding to a given point P is easily con- 





structed. We have only to erect a perpendicular to the trans- 
2 2 
verse axis at a distance a — EE WACO from the origin; that is, 
ae J a= mee Fe 
at a distance from the centre equal to the projection of 0 on 
the asymptote, and to join the projection of P on this line with 


the centre. The angle made by the joining line with the trans- 
y 


2 


Vee 


verse axis is ¢, for its tangent is clearly 


EXAMPLES. 
2 2 
(1) Find the length of the arc of the hyperbola fe ee 1, 


measured from the vertex to the point whose ordinate is 2. 
Ans. 2.194; 


(2) Show that ae tan¢ Ad is the distance from the centre to 
the normal at P. 


(3) Show that the limiting value approached by the difference 
between the arc and the portion of the asymptote cut off by a 
perpendicular upon it from P, as P recedes indefinitely from 


Cuap. XVI.] ELLIPTIC INTEGRALS. 263 


2 Gey \e 
ferred to as the difference between the length of the infinite arc 
of the hyperbola and the length of the asymptote. 
Show that in example (1) this difference is equal to 2.803. 


2 
the origin, is aeH € 4 a ree 5): This is generally re- 
e © 


The Pendulum. 


203. We have seen in Art. 176 that if a pendulum starts 
from rest at a point of its arc whose distance above the lowest 
point is %, the time required in rising from the lowest point to 
a point whose distance above the lowest point is y, is 





— Co) = 
; ld a 
= ca sa 2 em Age [1] 
Ng ° V1i—# sin?¢ Ng 
where k= a and sing = 


In the figure let A be the lowest point of the arc, B the 





highest point reached by the pendulum, and P the point 
reached at the expiration of the time ¢. Call AOB a, and 
AOP 0. 
Then “= 1—cosa, and J = V4 (1— cosa) =sin= =k. 
a 2a 2 
Consequently the modulus of the Elliptic Integral in [1] is 


264 INTEGRAL CALCULUS. (ART. 203. 


the sine of one-fourth the angle through which the pendulum 
swings. 





a 
SA eg peareey yo 6 
and = V4(1—cos@) =sin-, 
2a 2 
y fe 
2a 2 
and sind = eb i ies 82 . 





and therefore the sine of the amplitude of the Elliptic Integral 

in [1] is easily computed when the angle through which the 

pendulum has risen is given. When 6=a, sind=1, and 

b= x so that the time of a half-oscillation is a2 P(sin$, 5) 
f 

a confirmation of [7], Art. 176. The construction indicated 





in the figure gives the angle ¢, corresponding to any given are 
ase Or 


A ig cos AO'?E®, 


1 
2 Yo 





and \i=vi ~ cos AO) = sino & = sin A0Q. 
Yo 
Therefore ACQ = ¢. 


CHap. XVI. ] 


It is very easy to express the angle @ in terms of ¢. 


We have 


hence 


and 


then 


and 


ELLIPTIC INTEGRALS. 


t= ar(sins, $); 
g 2 
tf= P( sin’, 6) 

a 2 


Bee ethan NE mod sin” - 
2 2 a 2 
ieee rciny t Ig mod sin” : 
2 \ a 2 


265 


sind = 2sin“sn (2) dn (1/2) gone sins) 
2 a a \ 2 


EXAMPLES. 


1) A pendulum swings through an angle of 180°; required 
p = g g q 


the time of oscillation. 


Ans. 3.708 A 
g 


(2) Compare the times required by the pendulum in Ex. (1) 
to descend through the first 30°, the second 30°, and the third 
30° of its are respectively. 


Ans. 1.028 23 0.446 5 0.380 ae 
g g g 


(3) The time of vibration of a pendulum swinging in an are 
of 72° is observed to be 2 seconds; how long does it take it to 
fall through an are of 5°, beginning at a point 20° from the 


highest point of the arc of swing? 


Ans. 0.095 seconds. 


266 INTEGRAL CALCULUS. (ART. 203. 


(4) A pendulum for which Ne has been determined, and is 


equal to 4, vibrates through an arc of 180°; through what are 
does it rise in the first half-second after it has passed its lowest 
point? in the first ¢ of a second? Ans. 09° <0 


(5) It has been shown in Art. 176 that if y>2a the 
pendulum will make complete revolutions, and that the time 
required to pass from the lowest point to any point whose 
distance above the lowest point is y, is 


[2 ae ds | 2 
Sah —F (k, ) 
NV Gyo Ji — ke? sin’ d \ GY ; 








5 1A 
where on baat - and sing = NER 
Yo 


Show that in this case ¢ = a and that sin - Mh SD t lay . 
z 2 aN 2 


Notr.—In working with a pendulum it is often about as 
easy to compute F(k, ¢) by developing by the binomial 
theorem and integrating two or three terms, as to use a table 
of Elliptic Integrals. 


% dd 
ye have FI 8) ras 
—rh Sin 


(1 — k? sin’) Ve lt deisin’g +55 oi sin' p+: 





Se dd iy aie 
and F(k, $)= { vine Pasay | ae sing cos ¢) 


ae ai sin? cos@ + rr (¢ — sin d cos¢) «+++ 


Cuap. XVII. ] THEORY OF FUNCTIONS. 267 


CHAPTER XVII. 
INTRODUCTION TO THE THEORY OF FUNCTIONS. 


204. A function having but a single value for any given 
value, real or imaginary, of the variable is called a single-valued 
function. Rational Algebraic Functions, Exponential Func- 
tions, the direct Trigonometric Functions, and the Elliptic 
Functions are single-valued. 

A function which has in general two or more values for any 
given value of the variable is called a multiple-valwed function. 
Irrational Algebraic Functions, Logarithmic Functions, the 
inverse or anti-Trigonometric Functions, and the Elliptic In- 
tegrals, are multiple-valued. 


205. In Chapter II. we have explained the customary graph- 
ical method of representing an imaginary by the position of a 
point in a plane, the rectangular codrdinates of the point being 
the real term and the real coefficient of the pure imaginary term 
of the imaginary in question. 

In the ordinary treatment of the Theory of Functions this 
method of representation is of the greatest service, and enables 
us to bring the study of functions of imaginary variables within 
the province of Pure Geometry, and to give it great definiteness 
and precision. 

For the sake of brevity we shall in future use the symbol 7 
for V—1 and cis for cos¢+ V—I1sind, so that we shall 
write our typical imaginary as 7+yi or as rcis¢, instead of 
using the longer forms «+ yV—1, and r(cos¢ + V—Il1sing). 

We shall also use the name complex quantity for an imaginary 
of the typical form when it is necessary to distinguish it from 
a@ pure imaginary. 


268 INTEGRAL CALCULUS. TART. 206. 


206. A complex variable z=x-+ yi is said to vary continu- 
ously when it varies in such a manner that the path traced by 
the point (#,y) representing it is a continuous line. 





; 


Thus if z changes from the value a to the value f, so that 
the point representing it traces any of the four lines in the 
figure, 2 varies continuously. 

It will be seen that a variable can pass from the first to the 
second of two given values, real or imaginary, by any one of 
an infinite number of different paths without discontinuity if the 
variable in question is not restricted to real values; while a real 
variable can change continuously from one given value to another 
in but one way, since the point representing it is confined in its 
motion to the aais of reals. ; 


207. A single-valued function w of a complex variable z is 
called a continuous function if the point representing it traces 
a continuous path whenever the point representing z traces a 
continuous path. 

A multiple-valued function of z is continuous if each of the n 
points representing values corresponding to a value of z traces 
a continuous path whenever z traces a continuous path. These 
n paths are in general distinct, but two or more of them may 
intersect, a point of intersection corresponding to a value of z 
for which two or more of the n values of w, usually distinct, 
happen to coincide. Such a value of z is sometimes called a 


Cuar. XVII.] THEORY OF FUNCTIONS. 269 


critical value, and the consideration of critical values plays an 
important part in the Theory of Functions. 

In studying a multiple-valued function we may confine our 
attention to any one of its n values, and except for the possible 
presence of critical points this value may be treated just as we 
treat a single-valued function. 

In representing graphically the changes produced in a func- 
tion w by changing the variable z on which it depends, it is 
customary to avoid confusion by using separate sets of axes for 
w and z. 


208. If we use the word function in its widest sense, 
w=u-+vi will be a function of a complex variable z= a+ yi, 
if vw and v are any given functions of wand y. For example, 

wi, 6y, @+y’, w—yi, vw —y + 2ayi, het aha 
Ver + y+ 4 
may all be regarded as functions of z. 

We have seen in Chapter II., Arts. 86-42, that with this defi- 
nition of function the derivative with respect to z of a function 
of w is in general indeterminate; but that there are various 
functions of z, for instance, 2”, logz, e7, sinz, where the deriva- 
tive is not indeterminate. We are now ready to investigate 
more in detail the general question of the existence of a deter- 
minate derivative of a function of a complex variable. 

Let w=u-+vvi be a function of 2; uw and v, which are real, 
being functions of x and y. 

Starting with the value % = 2+ Y%? of z and the correspond- 
ing value w=uU+%t of w, let us change z by giving to x 
increment Aw without changing y. 





Let A,uw and A,v be the corresponding increments of u and 
vy; and z, and w, the new values of z and w. 


270 INTEGRAL CALCULUS. [ART. 208. 


We have Z=Z4+-Aw, w,=w+A,u+ idA,r. 


W,—W, A,u . tA,v 


Then os ; 
21 — 2p Aw Ax 








and the derivative of w with respect to z under the given cir- 
cumstances is 


limit | w,—w . 
aee0| poe |= Diu + Dav. [1] 


my — & 





If, however, starting with the same value 2 of z, we change 
z by giving y the increment Ay without changing 2, we have 


2 =X + (Yo + Ay)t = % + tAy, 
W, = Uy + AU + (VU +A,V)i= Wy +A, U + 7A,2, 


W,—Wy A,u , tA,v 
S252 ma 9 
zy — & vay way 





and 





limit E — Uy 


REN |= Div = iDy [2] 


ay — &% 
and this is the derivative of w with respect to 2 when we change 
y and do not change @. 

Comparing [1] with [2], we sce that if we start with a given 
value of z, and change z in the two different ways just con- 
sidered, the limits of the ratios of the corresponding changes in 
w to the changes in z need not be the same. Indeed, the two 


values for - given in [1] and [2] will not be the same unless 
dz 
w=u+vi is such a function of z=x+ yi that 


D,u=D,v and Du=—D,v. [3] 


Cuap. XVIL] THEORY OF FUNCTIONS. 271 


We shall now show that if w is such a function of z that 


equations [3] are satisfied, fone el will be the same if we 
Az 


start with a given value z% of z, no mattter in what manner z 
may change; that is, no matter in what direction the point 
representing z may be supposed to move; or, in other words, 


no matter what may be the value of ree bea | 


We have in general, since w is a function of the two variables 
x and y, 
Aw =(D,u+iD,v) Ax+(D,u+iD,v) Ay +e, 


where « is an infinitesimal of higher order than Aw or Ay. 
(1., Art. 198.) 
Az= Awv+iAy. 


Aw __ D,u.Ax+iD,v. Ay +iD,v.Au+D,u.Ay+e 





Hence / 
Az Aw +7Ay 
Duw+iDd,v- AL + iD, v+D, u- es 
— A a 
ee ped 
ie Ax 
ane limit |; Aw _ dw 
Az=0)| Az dz 
limit | Ay limit | Ay 
D,u+iDd,v. one ie tte Dy a0) ao) 
Ti . limit | Ay ‘ 
ur es 0 a. 4 [4] 


a value involving coe ey and therefore dependent upon 


the direction in which z is made to move. 
If, however, [3] is satisfied, [4] reduces to 


dw 
= D,u—iD,»v, 5 
a u—iD,v [5] 


and the derivative of w is independent of ph | 


272 INTEGRAL CALCULUS. (ArT. 209. 


A function which satisfies equations [3], and which, there- 
fore, has a derivation whose value depends only upon the value 
of the independent variable, and not upon the direction in which 
the point representing the variable is supposed to move, is called 
by some writers a monogenic function, by others a function which 
has a derivative; still others refuse to dignify with the name of 
function ary other than monogenic functions. 


209. Any function involving z as a whole, that is, any func- 
tion which can be formed by performing an operation or a 
series of operations upon z as a whole, without introducing x 
and y except as they occur in z, is a monogenic function of z. 


For if w=fz=f(x+yt), 
where fz can be formed by operating upon z as a whole, 
De fie. and Dw= Fe 
therefore 1D,w=D,w, or iD,(u+vi)=D,(u+);3 
_ whence Db = I, and D,u=— D,v; 


and [3], Art. 208, is satisfied. Consequently w is monogenic. 
This accounts for the results of Arts. 38-42. 
If w is a multiple-valued function of z, there may be several 


! dw 
different values of ae? corresponding to the same value of z; 
z 


but if w is monogenic, each of these values depends only upon 2, 
and not upon the way in which z is supposed to change. 

In future, unless something is said to the contrary, we shall 
give the name function only to monogenic functions. Thus we 
shall not call such expressions as x—yi, or #+y'+ 2ayi, 
functions of z. 


Conjugate Functions. 


210. Ifw and v are functions of w and y¥, satisfying equations 
[3], Art. 208, it is easy to prove that 


DZu+ Dju=0 and «= Dv + Dv=0. 


Cuap. XVII.] THEORY OF FUNCTIONS. 273 
For since Du Dv and D,0=— Du, 
we have D4 = 1) Dv and Dju=— D,D,»r, 


% 


DZv=—D,D,u and ey eo Ns : 


u and v are then solutions of Laplace’s equation, 
D2V + DgV=0. [1] 


Any two functions @ and wv of w and y, such that 
o (x,y) +iw(a,y) is a monogenic function of «+yi, are 
called conjugate functions ; and, by what has just been proved, 
each of a pair of conjugate functions is always a solution of 
Laplace’s Equation [1]. 

Thus 2—y*, 2ay; e*cosy, e*siny; stlog(a’+ 7’), tan 
are three pairs of conjugate functions, since 2° — y° + 2ayi 
=(x+yi)*, & cosy + ie siny=e**", Slog (a*+y*) +7 tan-t= 
= log (x + yi), and consequently, by Art. 209, are all monogenic. 
Therefore each of the six functions at the beginning of this 
paragraph is a solution of Laplace’s Equation [1]. 

It is clear that we can form pairs of conjugate functions at 
pleasure by merely forming functions of «+ yi and breaking 
them up into their real parts, and their pure imaginary parts ; 
that is, throwing them into the typical form uw + vi. 

If each of a pair of conjugate functions, ¢ and y, is written 
equal to a constant, the equations thus formed will represent a 
pair of curves which intersect at right angles. For let (a, 7) 
be a point of intersection of the curves 6 =a, Y=); the slopes 





; Thom Da 

of the two curves at (a, 7) are respectively — —*~, ——* by 
Do Dw 

I., Art. 202; and since D,¢=D,y and D,y~=— D,¢, the 


second slope is minus the reciprocal of the first, and the curves 
are perpendicular to each other at the point in question. 
Thus 2? — y? =a, 2ay= db, cut each other orthogonally ; as do 


274 INTEGRAL CALCULUS. (Arr. 211. 


also Slog (a@’?+y°)=a, tan 1 — 5; or, what amounts to the . 
a 
same thing, #+y=q x ba =b,. It must be observed, how- 
x 


ever, that 7+ y and Y are not conjugate functions, and that 
a 


in general the converse of our proposition does not hold. 

It may be easily proved that if ¢ and w are conjugate func- 
tions of x and y, and f and / are any second pair of conjugate 
functions of « and y, the new pair of functions formed by re- 
placing @ and y in ¢ and w by f and F respectively will be 
conjugate. 


Thus (e* cosy)? — (e*siny)*, 2e* cosy.e*siny, 
or, reducing, e*cos2y, e*sin2y, 


are conjugate functions ; 


blog [ (a®— y')? + (2ay)*], tan” ice 


ee 


y P 2 20 
or, reducing, log (a°+y’), ee . 
vy 


are conjugate. 
The properties of conjugate functions given in this article 
are of great importance in many branches of Mathematical 


Physics. 
EXAMPLE. 


Show that if #’ and y' are conjugate functions of a and y, 
v and y are conjugate functions of a! and y/’. 


Preservation of Angles. 


211. If w is a single-valued monogenic function of 2, and 
the point representing z traces two arcs intersecting at a given 
angle, the corresponding arcs traced by the point representing 
w will in general intersect at the same angle. 


Cap. XVII.] THEORY OF FUNCTIONS. 275 


For let z be the point of intersection of the curves in the z 
plane, and w, the corresponding point in the w plane. Let z, be 
a point on the first curve, and z, a point on the second; and let 





w, and w, be the corresponding points in the w figure. 

Let 7}, 72) $;, and s, be the moduli of 2; —%, 2, — 2%, Ww, — Wp, 
and w,—w, respectively, ¢;, ¢., %, and y their arguments ; 
then, since w is a monogenic function of z, we must have 


Bye eb ies, 00 Sapo Wao ay 
bmi p22 limit} —2— |, 
my — & m2 — % 


ee limit | C!S ¥4 | — pimit | S28 |, 
7, CIS dy 17 CIS ho 
whence, by Art. 25, 
limit a cis ( — #1) | = limit E Cis (Yo — $3) | ‘ 
i) us 


and since, when two imaginaries are equal, their moduli must 
be equal, and their arguments must be equal, unless the moduli 
are both zero or both infinite, 


limit (Ww. — u,) = limit (¢d2— 4) ; 


that is, the angle between the arcs in the w figure is equal to 
the angle between the corresponding arcs in the z figure; unless 


dw dw 
ear) = 0, or at =n. 
dz |z=z, dz |z=z, 


If w is a multiple-valued monogenic function of z, and if 
starting from any point 2, the point which represents z traces 


276 INTEGRAL CALCULUS. fArr. 212. 


out two curves intersecting at an angle a, each of the n points 
representing the corresponding values of w will trace out a pair 
of curves intersecting at the angle a; unless % is a point at 


Aer eLerres ; : 
which ae is zero or infinite. 
Z 


If, then, w is any monogenic function of z, and the point 
representing z is made to trace out any figure however complex, 
the point representing w will trace out a figure in which all the 
angles occurring in the z figure are preserved unchanged, except 
those having their vertices at points representing values of z 


: dw ; ) 
which make a zero or infinite. 
IZ 


This principle leads to the following working rule for trans- 
forming any given figure into another, in which the angles are 
preserved unchanged. 

Substitute 2’ and y' for x and y in the equations of the curves 
which compose the given figure, x’ and y’ being any pair of 
conjugate functions (Art. 210) of w and y, and the new 
equations thus obtained will represent a set of curves forming 
a second figure in which all the angles of the given figure are 
preserved unchanged, except those having their vertices at 
points at which D,#’ and D,y' are both zero, or at which one of 
them is infinite. ‘ 


For exniasple, eC—Y=A, (1) 
e+ y=b, (2) 


4 


are a pair of perpendicular right lines. Replace x by 2 — 7’ 
and y by 2ay, and we get 


oe —2ay—y=a, (3) 
e+ 2ay—y’=b, (4) 


a pair of hyperbolas that cut orthogonally. 


212. If w is a single-valued continuous function of 2, it is 
clear that if w) and w, are the values corresponding to % and 2,, 


Cuap. XVII.] THEORY OF FUNCTIONS. 217 


and the point z moves from % to 2, by two different paths, the 
corresponding paths traced by w will begin at wy) and end at w,, 
and consequently that if z describes any closed contour, -w also 
will describe a closed contour. 

If w is a double-valued function of z, since to each value of 
z there will correspond two values of w, it is conceivable that 
if w, and w,' are the values of w corresponding to z,, and 2 moves 
from % to 2, by two different paths, w may in one case move 
from wy to w,, and in the other case from wy to w,/. 

We can prove, however, that if the two paths traced by z do 
not enclose a critical point (Art. 207), and w is finite and con- 
tinuous for the portion of the plane considered, this will not 
take place, and that the two paths starting from w will 
terminate at the same point w. 

For as z traces the first path, each of the two points repre- 
senting the two values of w will trace a path, one starting at wy, 
and the other at w)', and unless the z path passes through a 
critical point, the two w paths will not intersect, but will be 
entirely separate and distinct, and will lead, one from w,' to w,, 
the other from w,' to w,!. 

If, now, the z path be gradually swung into a second position 
without changing its beginning or its end, since w is a continu- 
ous function, the two w paths will be gradually swung into new 
positions ; but, provided that the z path in its changing does not 
at any time pass through a critical point, the two w paths will 
at no time intersect, and consequently it will be impossible for 
the w points to pass over from one path to the other, and there- 
fore the point which starts at w, must always come out at w, 
and not at wy’. 

It follows readily from this reasoning that if z describes a 
closed contour not embracing a critical point, each of the w 
points will describe a closed contour, and these contours will 
not intersect. 

Of course, the proof given above holds for any multiple- 
valued function. 

In any portion of the plane, then, not containing critical 


278 INTEGRAL CALCULUS. (ArT. 213. 


points the separate values of a multiple-valued function may be 
separately considered, and may be regarded and treated as 
single-valued functions. 


213. That in the case of a double-valued function two paths 
in the z plane, including between them a critical point, but 
having the same beginning and the same end, may lead to 
different values of the function, is easily shown by an example. 

Let w=*Vz, and let z, starting with the value 1, move to the 
value —1 by the semi-circular path in the figure. That one of 





—l 0 tea 
Fig. 1. 


the corresponding values of w which starts with +1 will de- 
scribe the a ts shown in the figure, and will reach the 


point 1. cis* 3 or7z. If, however, 2 moves from +1 to —1 by 


Oe 


Hie, 2. 


the semi-circular path in the second figure, the value of w which 
starts with +1 will describe the quadrant shown in the second 


figure, and will reach the value 1.cis(—5), or —i. These 


two paths described by z, then, although beginning at the same 
point +1 and ending at the same point — 1, cause that value 
of the function which begins with +1 to reach two different 
values; and the two paths in question embrace the point z= 0, 
which is clearly a point at which the two values of w, ordinarily 
different, coincide ; that is, a critical point. | 


Cuap. XVII.] THEORY OF FUNCTIONS. 279 


It is easily seen that if 2, starting with the value +1, de- 
scribes a complete circumference about the origin, the value of 
w which starts from the point +1 will not describe a closed 
contour, but will move through a semi-circumference and end 
with the point 1.cist or —1. Now, by Art. 212 any path 


WIG aos 


described by z beginning with +1 and ending with — 1 and 
passing above the origin, since it can be deformed into the 
semi-circumference of Fig. 1 without passing through a critical 
point, will cause the value of w beginning with + 1 to end with 
+7; and any path described by z beginning with + 1 and end- 
ing with —1 and passing below the origin, since it can be 
deformed into the semi-circumference of Fig. 2 without passing 
through a critical point, will cause the value of w beginning 
with +1 to end with —7. Therefore any two paths described 
by z beginning with +1 and ending with —1 will, if they include 
the critical point z= 0 between them, lead to different values 
of w, provided that the same value of w is taken at the start. 


214. If wis a double-valued function of z, and z describes a 
closed contour about a single critical point, this contour may be 
deformed into a circle about the critical point, and a line lead- 
ing from the starting point to the circumference 


of the circle, without affecting the final value of 8 x 
w (Art. 212). Thus, in the figure, the two 

paths ABCDA, AB'C'D'B'A lead from the oy 
same initial to the same final value of w; and D 


this is true no matter how small the radius of O 
the circle B'C'D’. 


280 INTEGRAL CALCULUS. [ART. 214. 


Let 2% be the critical point, and let w) be the corresponding 
point in the w figure. As z moves from 2, towards %, the points 


Zo 





representing the corresponding values of w will start at w, and 
w,' and move towards wy, tracing distinct paths. 

If, now, z describes a circumference about 2%, and then 
returns along its original path to 2, the first value of w will 
either make a complete revolution about w) and return along 
the branch (1) to its initial value w,, or it will describe about 





Case II. 


w, a path ending with the branch (2) of the w curve, and move 
along that branch to the value w,’. 

In the first case, and in that case only, the value of w 
describes a closed contour when z describes a closed contour, 
and is practically a single-valued function. 

dw 


° ‘ : Oo. 5 . : 
If z% is a point at which — is neither zero nor infinite 
dz 


(v. Art. 211), when z describes about 2 a circle of infinitesimal 
radius, w will make about w, a complete revolution ; for since 
if two radii are drawn from 2, the curves corresponding to them 
will form at w) an angle equal to the angle between the radii, 
when a radius drawn to the moving point which is describing 
the circle about z revolves through an angle of 360°, the cor- 


Cuap. XVII.] THEORY OF FUNCTIONS. 281 


responding line joining w, with the moving point representing w 
will revolve through 360°, and we shall have what we have 
called Case I. 


If, then, we avoid the points at which a is zero or infinite, 
z 


we shall avoid all critical points that can vitiate the results 
obtained by treating our double-valued or multiple-valued func- 
tions as we treat single-valued functions. 

A critical point of such a character that when z describes a 
closed contour about it the corresponding path traced by any 
one of the values of zw is not closed, we shall call a branch point. 

When a function is finite, continuous, and single-valued for 
all values of z lying in a given portion of the z plane, or when 
if multiple-valued it is finite and continuous, and has no branch 
points in the portion of the plane in question, it is said to be 
holomorphic in that portion of the plane. 


Definite Integrals. 


eZ 
215. In the case of real variables, if jz.dz was defined in 


Art. 80 in effect as follows: 


{a dz= ae [ S20 (41 — 20) + f%1 (22 — 2) + fe (23 — 22) + °° 
; + fen1(Z-— Zn—1) |s [1] 


where 2), 25 2) «+» Z,-1 are values of z dividing the interval 
between % and Z into n parts, each of which is made to 
approach zero as its limit as 7 is indefinitely increased. 
Zz 
In other words, { is the line integral of fz (Art. 163) taken 


ad) 


along the straight line, joining z and Z if z and Z are repre- 
sented as in the Calculus of Imaginaries. 

It has proved that if /% is finite and continuous between 
& and Z, this integral depends merely upon the initial and final 
values of z, and is equal to FZ — F'% where Fz is the indefinite 


integral if. fz. dz. 


282 INTEGRAL CALCULUS. [ART. 215. 


If z is a complex variable, and passes from % to Z along any 
Z 
given path, we shall still define the definite integral { Sz. dz by 
70 
[1] where now 2, 2%, 23,.-.2,_, are points in the given path. 


Two important results follow immediately from this defini- 
tion : 


Zz Z 
ist. That f fe.de=— f fe. da, [2] 


if z traverses in each integral the same path connecting z and Z. 
Z 
2d. That the modulus of fz.dz is not greater than the 
= 
line-integral of the modulus of jz taken along the given path 
joining 2 and Z. 
If we let 


fe=W=U4+U, 2a=e+ yi, ub (2,y), and Oa 


Z 
then af jade ah ut vi) (dx + idy) 


= {$e ydw+ifu(z, ydr— fu, nay tif dayay, 
[3] 


each of the integrals in the last member being the line-integral 
of a real function of real variables, taken along the given path 
connecting % and Z. 

If the given path is changed, each of the integrals in the 
last member of [3] will in general change, and the value of 


Z 
fz. dz will change; and, since z may pass from 2 to Z by an 
a’ 2 
infinite number of different paths, we have no reason to expect 
i 
that | %.dz will in general be determinate. 
9 
We shall, however, prove that in a large and important class 


Z 
of cases f jz.dz is determinate, and depends for its value 
z 
0 


upon % and Z, and not at all upon the nature of the path 
traversed by z in going from % to Z. 


Cnap. XVII.] THEORY OF FUNCTIONS. 283 


216. If fz is holomorphic in a given portion of the plane, 


Sf ty) [1] 


if z describes any closed contour lying wholly within that 
portion of the plane. 
From [3], Art. 215, we have 


[fe de=fw.de= {ude +i f vde— vdy +i f udy, [2] 


the integral in each case being the line-integral around the 
closed contour in question. 

Since w= fz is holomorphic, u=¢(a2,y), and v=y (a, y), 
and D,u, D,u, D,v, and D,v are easily seen to be finite, con- 
tinuous, and single-valued in the portion of the plane considered. 
Therefore, by Art. 170, 


fe udx = f i Dyudady ; Hf vda = Hf; f Dyvdady ; 
( vdy = — { { Devcdady; ( udy = — { Dud«dy ; 


the integral in the first member of each equation being taken 

around the contour, and that in the second member being a 

surface-integral taken over the surface bounded by the contour. 
We have, then, from [2], 


{Re Pci f { (Dju+ D,v)dady + % { f (Djv—D,u)dady, [3] 
but D,uw = D,v, and Du = — D,v from [3], Art. 208. Therefore, 
[3] reduces to { f.& == 


From this result we get easily the very important fact that if 
Z 
fz is holomorphic in a given portion of the plane, | j.dz will 


have the same value for ail paths leading from 2 to Zz provided 
they lie wholly in the given part of the 


plane. For let z,aZ and 40Z be any 
two paths not intesecting between % Zo 
and Z. Then zaZbz is a closed con- a 


tour, and p 


284 INTEGRAL CALCULUS. [ART. 217. 


ih i Jz. dz (along %aZb%) 
Z %Z 
=f fz. dz (along 2aZ) + | jz. dz (along Zb%) = 0; 
zy Z 


zy Z 
but f fz. dz (along 26%) = — { fz.dz (along %bZ) 
Z 2 
? by Art. 215. 
Z Z 
Therefore, i fz. dz (along 2aZ) = { fz. dz (along 4bZ). 


a be If the paths z%aZ and 2 bZ inter- 
sect, a third path zcZ may be drawn 
not intersecting either of them, and 
by the proof, just given 


C 
Z eZ 
i Jfz.dz (along zaZ ) al Jfz.dz (along %cZ), 


Z Z 
ue fz.dz (along %bZ) =f fz.dz (along 2cZ) ; 
therefore, 


Z wZ 
Jf fz. dz (along %aZ) = | Jz. dz (along %bZ). 


217. If fz, while in other respects holomorphic in a given 
portion of the plane, becomes infinite for a value 7’ of z, then 


Hi jz.dz taken around a closed contour embracing 7’, while not 


zero, is, however, equal to the integral taken around any other 
closed path surrounding 7’. 

For let ABCD be any closed con- 
tour about 7. With 7 as a centre, 
and a radius e, describe a circumfer- 
ence, taking « so small that the cir- 

-cumference lies wholly within ABCD. 
Join the two contours by a line AA’. 
Then ABCDAA'D'C'BIA'A is a 
closed path within which jz is holo- 
morphic. 





Cuap. XVII.] THEORY OF FUNCTIONS. 285 


Therefore, 
ne fz.dz (along ABCDAA'D'C'B'A'A) =0, 


or fiw (along ABCDA) + (fede (along AA’) 


+ [x dz (along A'D'C'B'A') + ff. dz (along A’A)=0; 
but : 
fx .dz (along AA’) =— (fede along (A'A), 


fv coeraions A’ Dp!" BA’) = — {fz .dz (along A'B'C'D'A'). 


fz. dz (along ABCDA) = fied (along A'B'C'D'A'). 


218. That the integral of a function of z around a closed 
contour embracing a point at which the function is infinite is 
not necessarily zero is easily shown by an example. 





1 ea 
fez t being a given constant, is single-valued, con- 


tinuous, and finite throughout the whole of the plane except at 
1 





the point ¢, at which becomes infinite, without, however, 
ceasing to be single-valued. 


Let us take f ae around a circle whose centre is ¢, and 





whose radius is any arbitrarily chosen value «. If z is on the 
circumference of this circle 


z—t=e(cos¢+ ising) 
= ee”! by [5], Art. 31. 


zg=t+tee* 





and dz = ice*' dd. 


286 INTEGRAL CALCULUS. (Arr. 218. 


2 i 
Hence f ia (around abc) =f a 
z2—t 0 ceri 


From what has been proved in Art. 217, it follows that 





f ow around any closed contour embracing ¢ must also be 


equal to 277. 
As another example let us consider Ht ae, when fz is 
Z— 


supposed to be holomorphic in the portion of the plane con- 
sidered, and where the integral is to be taken around any closed 
contour embracing the point z= ¢. 
& is holomorphic except at the point z=t, where it 
Z— 
becomes infinite. The required integral is, then, equal to the 
integral around a circumference described from the point ¢ as 
a centre, with any given radius e, that is, by the reasoning just 
dz 


gah 


or pi\ 7 _ pt Qn 
f F'(t + ce") tee itd aM if F(t + ece*) dd; 
0 0 


ee? 


used in the case of to 








and in this expression « may be taken at pleasure. If now e is 
made infinitesimal ee* is infinitesimal, and since F% is continu- 
ous F(t + ee*) is equal to #t+ 7 where y is some infinitesimal, 
and F'(t + ee*) d¢ is equal to Ft.dé + 7.d¢. 

Now, by I. Art. 161, 


27 Qa 
f (Ft.dg +16) = { Ft.dé. 
0 0 
27 q QT 
Hence if F(t + ce*) dd =a) Ft.db=2aiFt: 
0 e/0 


and we get the important result that {== dz, taken around any 
are 

contour including the point z= ¢, is equal to 277. Ft. 

1 Fz 5 


From this we have Uy mene : 
2mrid z—t 


? 


Cuap. XVII.] THEORY OF FUNCTIONS. 287 


and we see that a holomorphic function is determined every- 
where inside a closed contour if its value is given at every point 
of the contour. 
If inthe forma =. (“«@ [1] 
QriJ z—t 





we change ¢t to ¢ + At, we get 


Ruma Fz.dz me MARI 8  Ma.da. At dz. At 
2rt i — RELY Cae heaAry 











cn) t—At et dem t) (2 teaAty,” 
whence 
limit | AF¢ | __ ais payee. limit 1 
BI OAEN 277 > | &6=0) (2 — OH) (2 —t— at) | 
A Rees zr abet [2] 
2m) (z—Tt)° 
and in like manner we get 
Fz.dz 
Ft= ; 5 
Qari rt (z _ Aka [ J 
! 
and in general pe eek Had San [4] 


Dai (2. t)ry 


each of the integrals in these formulas being taken around a 
closed contour lying wholly in that portion of the plane in which 
Fz is holomorphic, and enclosing the point z=. 


219. The integral of a holomorphic function along any given 
path is finite and determinate, for, by [38], Art. 215, it is equal 
_ to the sum of four line integrals, each of which is finite and 

determinate (Art. 166). 

Tf & series Wy + wy + Wot-++, where wo, Wy, Wess are holo- 
morphic functions of z, is convergent for all values of z in a 
certain portion of the plane, the integral of the series along any 
given path lying in that portion of the plane ts the series formed 
of the integrals of the terms of the given series along the path in 
question, and the new series is convergent. 


288 INTEGRAL CALCULUS. FART. 220. 


For let S = Wy + wy Ht We bees Wr Wag He 
= Wyo +W+W+:+u,+h, 


where R= Writ Wayot +o, 
and where by hypothesis A (ie) =v; (1) 


[Sade = f rwode + {wide + +f w,dz +f R. de 


for any given value of n. 
By the proposition at the beginning of this article, uf Sdz 


along the given path is finite and determinate, as are also 


frvde, ford, etc. 


If, now, n is indefinitely increased, 


{Sa = {mde+ w, dz + f wide + +e + int ( Raz. 


The modulus of f dz is not greater than the line-integral 


along the given path of the modulus of # (v. Art. 215). Asn 
increases each value of the modulus # approaches zero as its 
limit (1) ; consequently each element of the cylindrical surface 
representing the line-integral of the modulus of # (v. Art. 166) 


approaches zero, and oe Af Rdz= 0. 


Therefore, {Saez = { mt ae fora + mde + : [2] 


and, since the first member is finite and determinate, the second 
member is a convergent series. 


Taylor’s and Maclaurin’s Theorems. 


220. te leith Fica ea lee lear 


identically, if n is a positive integer, even when q is imaginary. 


Cuap. XVILI.] THEORY OF FUNCTIONS. 289 


If the modulus of g is less than 1, 





limits at 
n=o LT] = 
; a limit | 1 — q” 1 
Hence 1l+q+@+q+--= ——4. |= 4 1 
q+¢+¢q ttf Tan [1] 


even when gq is imaginary, provided that the modulus of gq is 
less than 1. 

Suppose, now, that everywhere within and on a certain cir- 
cumference described with the point z=a as a centre F% is 
holomorphic. Let z=¢ be any point within this circumference, 
and z= Z be a point on the circum- 
ference. Then the modulus of Z—a 








is the distance from a to Z, and the : 
modulus of ¢—a is the distance from 
a tot; 
hence mod(t—a)< mod(Z—a), 
; O 
and mod (7% ahs 1 
asi eat 1 
eee oa) Za i. 7 
Z—a 
1 t—a (t—a)? | (t—a)’ 
ae 1 be ALN TR ye 
ar Sy on ie (Zoey cy Cec 
Hence Bene 
Oa | t—a (t{—a)? eS (t— a)? + +--+, [2] 


eee a2 a)? (Z—a)® .(Z— a)? 
and the second member of [2] is a convergent series. 
Multiply [2] by a and the series will still be convergent 
270 


for each value of z which we have to consider; we get 








1 FZ 

2mri Z—t 
1 FZ FZ : 
pale te a Ft ae a) Sea [3] 


290 INTEGRAL CALCULUS. [ART. 220. 


Integrate now both members of [3] around the circumfer- 
ence, and we have 


FZ 1 FZ FZ 
a da= | (F* azt(r— {Gv 
Pai, Poe mal Goa he 





FZ 
(Za)? 


and, since each of the functions to be integrated is holomorphic 
on the contour around which the integral is taken, and the 
second member of [3] is convergent, each integral will be finite 
and determinate, and the second member of [4] will be con- 
vergent. 

Substituting in [4] the values obtained in Art. 218, [1], [2], 
[3], and [4], we have 


+(t—a)? retires Oe aren a [4] 


Ft= Fa+(t—a) Fa +29 Pg —+ tay Fa +.-- 
+ CaO" pg $a, [5] 
If the point z= a is at the origin, a= 0 and [5] becomes 
Fi= Fo+tPo+ Flot Sot, 6] 
which is Maclaurin’s Theorem. 


That [5] is merely a new form of Taylor’s Theorem is easily 
seen if we let t—a=h, whence t=a-+h, and [5] becomes 


h? 


F(a+th) = Fath F'a +5 “Fra a Sh nae a —— J ieee [7] 
[6] can, of course, be written 
z= Fo+2F'o+— = D1 (Pot SFM b [8] 


and [5] as 
z= Fa+(z—a) Fla+ a sik er a afiesue 
[9] 


Cuap. XVII.] THEORY OF FUNCTIONS. 291 


and we get the very important result that if a function of z is 
holomorphic within a circle whose centre is at the origin it may 
be developed by Maclaurin’s Theorem, and the development will 
hold, that is, the series will be convergent, for all values of z 
lying within the circle. 

If a function of z is holomorphic within a circle described 
from z=a as a centre it can be developed by Taylor’s Theorem 
into a series arranged according to powers of z—a, and the 
development will hold for all values of z lying within the circle. 

The question of the convergency of either Taylor’s or 
Maclaurin’s Series for the case when z lies on the circum- 
ference of the circle needs special investigation, and will not. 
be considered here. 

If the function which we wish to develop is single-valued, in 
drawing our circle of convergence we need avoid only those 
points at which the function becomes infinite; but if it is 
multiple-valued we must avoid also those at which its derivative 
is zero or infinite (v. Art. 214). 


221. We are now able to investigate from a new point of 
view the question of the convergence of the series obtained by 
Taylor’s and Maclaurin’s Theorems in I. Chap. IX. 

Let us begin with the Binomial Theorem, 


n n n— (2 erid yn—2 J,2 
(a) (ath)*=a"+ na tna hi? + see, [1] 
or, following the notation of [9], Art. 220, 


a= a" + na" "(Zz —a)+ OD) a-2¢z —ayvte--. [2] 

If n is a positive integer, 2” is holomorphic throughout the 
whole plane, and [2] holds for all values of z and a, and [1] 
for all values of a and h. 

If n is a negative integer, 2” is single-valued, and it is finite 
and continuous except for z=0, where 2” becomes infinite. 
[2] is, then, convergent for all values of z lying within a circle 
described with a as a centre and passing through the origin; 


292 INTEGRAL CALCULUS. TART. 221. 


that is, for all values of z, such that mod (z —a)< moda; and 
consequently [1] holds if modh < moda. 
If nis a fraction, 2" is multiple-valued, and our circle of 


ky US 
convergence must avoid the points at which re becomes zero 
z 


or infinite ; but as the origin is the only point of this character, 
the circle of convergence is the same as in the case last con- 
sidered, and [1] holds for all cases where mod h < moda. 

When « and / are real our results agree with those obtained 
in I. Art. 131. 


(dD) e? = e* tv’ — e* (cosy +7siny) ({4], Art. 31) 


% 
is single-valued and continuous, and becomes infinite only when 
w%=o. Maclaurin’s development for e* holds, then, for all 
finite values of z. 


(c) logz = log (7 cis?) = logr + gi (Art. 33) 
is finite and continuous throughout the whole plane. It is, 
however, multiple-valued, but its derivative E becomes infinite 

z 


only when z= 0, and does not become zero for any finite value 

of z. logz, then, can be developed into a convergent series, 

arranged according to powers of z — a, for all values of 2 within 

a circle having the centre a@ and passing through the origin ; 

that is, for all cases where mod (z — a) < moda. 
If z—a=h, we get 

h? h* ht 


h 
log’ +h) = loga + —— — + — — = 
Oe oe fy Qi 20 OS 0, eee 


vee, [3] 


[3] holding for all cases where mod h < mod a. 
Ifa=1 andh=z, we get 
2 


; y2 yf 
log (1+z)= TF Oita eae [4] 


od 
] 
which holds for all values of 2 where mod z < 1. 


(d) sing =sin (7+ yi) = sine. F"" 4 jeose.- 2 = 2", 





Cuap. XVII. ] THEORY OF FUNCTIONS. 293 


is Si Ch 
—isine-- 
2 
(v. [3] and [4], Art. 35) 
are single-valued, and are finite apd continuous throughout the 
plane. ‘Therefore, Maclaurin’s developments for sinz and cosz 
hold for all values of z. 


sin z 
(e) tanz= , and secz = 
COS Z 


and cosz=cos(#-+ yi) = cosa: 





’ 


Ce." 
2 








, are single-valued and 
COS Z 


continuous, and become infinite only when cosz=0; that is, 


T ; poe 
when z = 55 Therefore, Maclaurin’s developments for tanz and 


secz (I. Art. 138), hold for every value of z whose modulus is 


less than <- 
2 


COSZ 1 : : 
(f) ctnz = —, and csez = —— become infinite when z = 0, 
sinz sin z 


and cannot be developed by Maclaurin’s Theorem. 





(g) sin-+z is-finite and continuous throughout the plane ; it 


becomes 





is, however, multiple-valued, and its derivative ny pine 
infinite when z=1, and whenz=—1. Therefore, the develop- 
ment for sin-’z (I. Art. 135 [2]), holds for any value of z 


whose modulus is less than 1. 


(h) tan~'z is finite and continuous throughout the plane ; it 





becomes infinite 


is multiple-valued, and its derivative —— 
when z =i, and when z=—?. ‘Therefore, the development for 
tan-'z (I. Art. 135 [1]), holds if modz< modi; that is, if 
modz<l. 

EXAMPLES. 

z 
1+2 
iiepeea rt. too, Bx. 1, holds if. modz< 1. 

(2) Show that the development of log (1+), given in I. 
Art. 186, Ex. 2, holds if modz<7z. 


(1) Show that the development of + log (1+ 2), given 





(3) Obtain the following developments, and find for what 
real values of x they hold good: 


294 




















INTEGRAL CALCULUS. (Arr. 221. 
1 Shae 
a) log (a+ Va? +a? Oe th 5 a 
Ae Re eal sa) a a ae a as 
4 
(dD) (e* +e Ae ao(4m +n(3n— ao 
Sa ry ie ' 2 a° 4 at 
(c) .e*.cos” mccmcamargs <1 
6 
(d) bees =(1-5 4a! a 
eid ts 6! 
x . _ Vase Aad 2 9 x 
(e) elog(it2)  =04+5 42545"... 
6 8 
(f) tan*x More es Uda a: 
o Q 
7 nat 
(g) (1+2e@+ 3807)? =1—x%+22 ot 
ay), no 

h) en '* = 1+ 2 ay sae 
( ) +a+ > 5 
1+ oe Rig 

log | ——— = ? pie Bes Fy 
(i) lime) Be. ) 

25, 2.17, 2.81 5 een 
()) tanv= eT oF = = oi42 71 v Br 11! Dwele’ 
(Oe PO 2x0 

3 45 945 4725 93555 

(7) log tan( +2)=log tant +20+— 8 oe 
7! 

seo a 3 a4 8a 83 x 56 a" 
(ety = lh aioe an on oa 
3 a 

(n) e ied Gare, (ae 4 Se + Se we 

1.9.39? . Lae 

O Ver Sin? '2)? == 2 | ee eee el 

Chas y (#+55 +35 3t357 4b ) 
1 2.) Zoe. ene Dari es aaa 
OF mrrewr marin amt 
1+2 3 4¥, war Bula ‘Shares 
(q) 6 —5a+2? 9 4 ee ae cae set e 




















Cuap. XVII. ] THEORY OF FUNCTIONS. 


Answers. 
(a) —a<ax<a; 


(b) —w<u<cowifn>0, —5<e<j ifn<0; 


(c) —wn<u<cn; 
(e) —1l<a<l1; 
Pa, v3. 
3 


ae <1; 
(kK) —3r7<u<cr; 
(m) —-x<x<o; 


(0) —2<%7<2; 
(q) —2<%<2. 


(d) —wo< rca; 


7 + 


(Ay HV eee 1; 


(py Via eo V2 


295 


296 INTEGRAL CALCULUS. [ART. 222. 


CAVE ON Vera Te 
KEY TO THE SOLUTION OF DIFFERENTIAL EQUATIONS. 


222. In this chapter an analytical key leads to a set of con- 
cise, practical rules, embodying most of the ordinary methods 
employed in solving differential equations ; and the attempt has 
been made to render these rules so explicit that they may be 
understood and applied by any one who has mastered the Inte- 
gral Calculus proper. 

The key is based upon ‘ Boole’s Dittoronnas Equations ” 
(London: Macmillan & Co.), to which or to ‘* Forsyth’s Differ- 
ential Equations’? (London: Macmillan & Co.), we refer the 
student who wishes to become familiar with the theoretical 
considerations upon which the working rules are based. 


223. A differential equation is an expressed relation involy- 
ing derivatives with or without the primitive variables from 
which they are derived. 

For example: 





(14+e)y+(1—y)et=0, (1) 
dy 
Hie a = 1 2 
re ay=%+ (2) 
ot —y ave ip y= 0, (3) 
dy \* dy 
se 4 
Gs v(y cB ot), Ce 
d‘y Cavin ok y¥_ofy 
—~—2—~+2 — rig | dD 
dac* dav? ir da? “dete ‘ () 


Cuap. XVIII.] DIFFERENTIAL EQUATIONS. KEY. 297 





sin?'a 4 + sina cosa’Y — y =a — sing, (6) 
da? da 

gry 
«(1 — «w)°—_+¥ —2y=0 r, 
2 6 
cat (ia )y=0 (8) 
da- a 
D?2—aD72z=0, (9) 


are differential equations. 

The order of a differential equation is the same as that of the 
derivative of highest order which appears in the equation. 

_ Equations (1), (2), (8), and (4) are of the first order; (6), 
(7), (8), and (9) of the second order; and (5) of the fourth 
order. 

The degree of a differential equation is the same as the power 
to which the derivative of highest order in the equation is 
raised, that derivative being supposed to enter into the equation 
in a rational form. 

Equations (1), (2), (8), (5), (6), (7), (8), and (9) are all 
of the first degree ; (4) is of the third degree. 

A differential equation is linear when it would be of the first 
degree if the dependent variable and all its derivatives were 
regarded as unknown quantities. 

Equations (2), (5), (6), (7), (8), and (9) are linear. 

The equation not containing differentials or derivatives, and 
expressing the most general relation between the primitive vari- 
ables consistent with the given differential equation, is called 
its general solution or complete primitive. A general solution 
will always contain arbitrary constants or arbitrary functions. 

The differential equation is formed from the complete primi- 
tive by direct differentiation, or by differentiation and the 
subsequent elimination of constants or functions between the 
primitive and the derived equations. 

If it has been formed by differentiation only without sub- 
sequent elimination or reduction, the differential equation is 
said to be exact. 


298 INTEGRAL CALCULUS. (ART. 224. 


A singular solution of a differential equation is a relation 
between the primitive variables which satisfies the differential 
equation by means of the values which it gives to the deriva- 
tives, but which cannot be obtained from the complete primitive 
by giving particular values to the arbitrary constants. 


224. We shall illustrate the use of the key by solving equa- 
tions (1), (2),:(3), (4), (5), (6), (7), (8), and (9) of Art. 223 
by its aid. 


(1) (1+2)y+(1—y)o!=0, or (1+2)ydx+ (1—y)ady=0. 


Beginning at the beginning of the key, we see that we have a 
single equation, and hence look under I., p. 310; it involves 
ordinary derivatives: we are then directed to II., p. 310; it 
contains two variables: we go to III., p. 310; it is of the first 
order, IV., p. 310, and of the first degree, V., p. 310. 

It is reducible to the form 

lee vy pari Loy dy = 0, 
a y 


which comes under Xdaw+ Ydy=0. 


Hence we turn to (1), p. 314, and there find the specific direc- 
tions for its solution. Integrating each term separately, we get 
logz+x+logy—y=c, or log (#y)+x—y=Cc, 

the required primitive equation. 
dy 
2 e~—ay=ut+l. 
(2) PTS men 


Beginning again at the beginning of the key, we are directed 
through I., Il., III., IV., to V., p. 310. Looking under V., 
we see that it will come under either the third or the fourth 
head. Let us try the fourth; we are referred to (4), p. 314, 
for specific directions. 

Obeying instructions, the work is as follows: 


dy 
y—*+ — ay =0, 
dla y 


Cuap. XVIII.] DIFFERENTIAL EQUATIONS. KEY. 299 


xdy — aydx = 0, 


dy ade _ 1" 
aes. iS 
logy —alogx=c, / 
i| 
log+=c; \ 
\ | 
UES oc ‘ 
Lhd 
=. Cr", (1) 
dy aCaxt 4 ga’, 
daz 
Substitute in the given equation, 
aCax* + nde —aCx*=x+1, 
eo 


TC (a+1)=0, 





Ae ied: 
gett 


CU ee ees (I 


(a—1)a** ax 





Substitute this value for C in (1), and we get 
1 
bran Cha" a re od ); 


a a—l 





the required primitive. 


(3) oY _ytaVe—¥=0. 
dx 

Beginning at the beginning of the key, we are directed 
through I., II., IIlI., IV., to V., page 310. Looking under V., 
we find that our equation does not come under any of the 
special forms there given. We are consequently driven to 
obtaining a solution in the form of a series, and for specific 
instructions we are referred to (13), page 316. Obeying 
these, our process is the following : 


300 INTEGRAL CALCULUS. [ART. 224. 














oo at _ vary, CUo = We _ fag? — yd, 
ade.) 04 LT 

? PRUE sa may a 2 ee 
Oo —y —-Vae — a, ; Yo — yy — i nt yeh, 
dx 2 diy Hs 
ay BY a/'x2 2 a? Yo 3 Yo Jn? at 

= FH a — 5 ae — : 

dec’ “y ie da,° Xp TOO ee 
ies NS pm rane i 4 
Oty 4+ive—y, : 0 — yy + — Ving — yo, 
da x lay XL 
RY LOY Wy aie 2 PY) 5 Yo v/ 20,2 2 
da oa dx a 
dé 6 dé 
oy Va? — y, Ho y — 2 Vx? — y4?, 
cla dx, Lo 

7 i SOUNBHE 5 7 ae 
mate t Va | GAS 
dx HX d. 0 Xo 


and the general value of y is 


pacha s 2 
Y= wot (ea) (8 Vagye)— SS (0 + Va a) 


Xs 
ee (“Vara =¥) 











3! 
x2 — X)* Oe Se Ee x — a)? /5 
+O (et Nout) to 
(hey ale be 
Sie rh Yoo Na! — Yo cee 


This result can be very greatly simplified by breaking up the 
series ; we have 


fib al i )e (Tay _ ay. 


+ n@=A( aa), Ga) eee 








4} 6! 


— Vai Ht ((@—2) — Goar 5 Gay Con...) 


CHap. XVIII.] DIFFERENTIAL EQUATIONS. KEY. 301 
re, aaemmea Wh eee x — 2,)* v—X)° 
— Vary? —Y aces ; ( — %)— Nosema o) “fs oe 51 oy 
0 . 


3! 


ye 
or 
m2 2 7 
y= >| cos (& — x) — Nae? = 10? sin (% — %) 

X% X% 

ae TR 
= | cos (X% — X%) — vr pes ae. - sin (oa) | 
X Ly 


Y0 is entirely arbitrary ; call it sina, then 
Xo 


y = x[sina cos(# — 2%) —cosa sin(# — %) |= wsinfa—(# — %) |, 


y =a sin(c— a), where c is any constant. 


dy\° dy 
4 —/\ =y7! v— }- 
(4) a) (y+ aot) 
Beginning at the beginning of the key, we are directed 
through I., II., I1I., 1V., to VI., page 311. Looking under VI. 


we see that the equation is of the first degree in 7; we are 
referred to (17), page 318, for our specific instructions. 


Obeying these, we first replace a by p; the equation becomes 
x 


p=y (y+2p). 
Differentiate relatively to y, and we get 
3p — 4y5 (y + ap) + 2y' + ay DP. 
dy dy 
Eliminate 2, 


dp,” y\ dp 
32? P — 4P 4 27! eat A Ne 
ee re y +(p alae 


or 2pe+y¥° dp_5(2pi+y') _ 9. 
prernay y 


302 INTEGRAL CALCULUS. [ART. 224, 


Striking out the factor 2p°+ y’, we have 
1 dp 2 
pdy y 
a differential equation of the first order and degree in which 


the variables are separated, and which therefore can be solved 
by (1), page 314. 


Its solution is log p — logy° = C, 
or es Baie 
y 


Eliminating p between this and the given equation, and re- 
ducing, we have cy(#—c*)=1, as our required solution. 
(5) Ty Oy od yey ees (1) 
dat dx da? dx 


Beginning at the beginning of the key, we are directed 
through I., Il., III., VII., to (22) (a), page 319, for our 
specific directions. 

We see at once that y = 1 is a particular solution. 

Obeying directions, we have now to solve 


dy By) Saye ay 
~~ 2 + 28 — 2+ y= 0 by 


Let y =e”, and we have 
m — 2m + 2m?—2m+1=—0, 
as our auxiliary algebraic equation in m. Its roots are 
1 iepeg Beat as © ese ETT. 
The solution of (2) is then 
y=(A+ Ba) e& + Ceosx+ Dsinz, 


and of (1) is 
y= (A+ Be) e* + Ccoosa+ Dsing +1. 


Cuap. XVIII.} DIFFERENTIAL EQUATIONS. KEY. 308 


2 
(6) sinter 4 + sine cose! — y = x — sina, (1) 
Beginning at the beginning of the key, we are directed 
through I., I]., III., VII., to (24), page 321, for our specific 
instructions. 
Dividing through by sin’a, the equation becomes 


ay dy 2 2 

pa ee es ome . eaettal — ph y 

dae nee cse’x. ¥ =X CSC LX — CSCX (2) 
y = ctn@ is found by inspection to be a solution of 

ay dy 2 a — e 


(2) can then be solved by (24) (a). 
Substitute y=zctna in (2), and it becomes 


ei har dz 
tan «— ctn?a — 2 ese?a) — = waese’?x — csex 
aa Ve : 


az 


dz 
or pia 7 bane + Bece CSC@) Re eancC spe ta seca. (3) 
dx 


Referring to (25), page 323, and obeying instructions, we 


let = S, and (3) becomes 
x 


dz! 
— —(tanw-+ secxcscx) 2’ = xseca cscx — seca, 


dx 


a linear differential equation of the first order ia z'. whose solu- 
tion by (4), page 314, is 


z' = A tanx secu — wsec’a + tan x seca (log tan 5 — log sin 2) ; 
but 2/= @, whence integrating, we have 
da 
z2=B+ Aseca —xtanae — (1+ secz) log (1+ cosa), 


and 
y = Acscx+ Betnx —x«—(escx+ctnz) log (1 + cosz). 


304 INTEGRAL CALCULUS. [ArrT. 224, 


Beit ad? 
(7) w(1—2)? 2 —2y=0. 


Beginning at the beginning of the key, we are directed 
through I., I., Ill., VII., to (24), page 321, for our specific 
instructions. . 

Let us try the method of (24) (e), page 323. 

Assume y= 2a, 2", and substitute in the given equation ; 
we have 

=[m (m — 1) a,,2"-' — 2m (m—1) 4,0" 
+m(m—1)a,,0"t! — 2a,,0"] = 0. 


Writing the coefficient of 2” in this sum equal to zero, we 
have 3 


m(m +1) daa — 2[m(m—1) +1 Ja, + (m —1)(m — 2) a,,_1=9, 
and we wish to choose the simplest set of values that will 
satisfy this relation. 
Substituting m=0, m=—1, m=-— 2, etc., in this relation, 
we find 
Ob l= Cos a => O_ 19 A_3 ae | A_95 ooo, 
Hence if we take a = VU, it follows that 
Aj = A_»=A_3+++ =D, 


and no negative powers of a will occur in our particular 
solution. 
Substituting now m=1, m= 2, m =3, etc., we have 


A = Ag = Ag—Ag—-rs. 


Taking a, = 1, we get as our required particular solution of the 
given equation 
Y =o + + oF + att oe, 


This can be written in finite form, since we know that 


Lee eee ie eee 
1—2z 
x 


1—2 





Hence y= 


is a particular solution. 


CHap. XVIII.] DIFFERENTIAL EQUATIONS. KEY. 305 


Turning now to (24) (a), page 321, we find 








yaya te (i tet za, lose) 
a? y 2? 
(8 “ 1——\y=— 0. 
(8) Fan ay 


Beginning at the beginning of the key, we are directed 
through I., II., ILI., VII., to (24), page 321, for our specific 
instructions. Let us try again the method (24) (e), page 323. 

Assume ¥ = 3a,,7”", and substitute in the given equation, 


Sm (m—1)a,,2"-? + a,,0" — 24,0" 7] = 0. 
The terms containing x” are 
(m +2) (M+1) Anse@” + An L” — 2 Am i9%™ § 
writing the sum of the coefficients equal to zero, we have 
Mm (M+ 3) Anis + A, = 0. (1) 


Letting m = 0 and m = —3, we get a = 0 and a_;=0; and all 
terms of y involving even negative powers of & disappear, as do 
all terms involving odd negative powers, except the — Ist. 

OL 




















In general ONG Se een Cos. alana 2 
x a m (m+ 1) (2) 
From this we get 
a 1 : 
“tere Seca if we take a, =1, 
Ac = Ay hs 1 
f 2.4:5.7 517 
Ay 1 
Ag = = =— —-, 
2.4.5.6.7.9 Tio 
Ay 1 
Ay = —— ——___—e 
2.4.5.6.7.8.9.11 viel 
é 4 6 8 10 
Hence Nena oe s a ad 


8. 315° 517719 Ol 
is a particular solution of the given equation. This can be 
thrown into finite form without much labor. 


306 INTEGRAL CALCULUS. (ART. 224, 


x a! a gl! 
ee -—— eo “¢ 


We have 9 0 3 751520817 ard One 


=a sing ; 
whence xy = sine —x cosa, 
Levee 
and = — (sin® —# cose). 


By going back to (2), and using odd values of m, we get 
anotber solution of our given equation, namely, 
xe a a 
214 4 4!6 618 





ye 
Y= . or 2 ’ 
which can be reduced to 
y= : (cosx-+ a sina). 
wv 


Hence our complete solution is 


y= [A(cosx +a sina) + B(sine —2cos2)], 
c 

3 4 

ae y = eee + sin (a — ”) b 
x 

B 
f we let —= tanec. 
if w A 


(9) D2Z2—-—av D72=9. 


Beginning at the beginning of the key, we are directed 
through I. and IX. to (45), p. 331, for our specific instruc- 
tions. 


Cuap. XVIII.] DIFFERENTIAL EQUATIONS. KEY. 307 


Obeying these, our work is as follows: 


dy’? — a’ dx’ = 0, 

dy —adx =0, (1) 

dy +adxa =0, (2) 
dpdy — a? dpdx = 0. (3) 


Combining (1) and (3), we get 
dpdy — adqdy = 0, 


or dp —adq=0. (4) 
(1) gives y—ar=a. - 
(4) gives p—aq=B. 
(2) and (3) give us, in the same way, 
Y + Ae = ay, 
pt+aqg=fi; 
and our two first integrals are ; 
p—ag=fi(y— aa), ah CO} 
pt+aq=frly + ae), (6) 


jf, and f, denoting arbitrary functions. 


Determining p and q, from (5) and (6), 


p=3lh(y tae) +f: (y—a2)], 
g=5-L iy +42) —fi(y—ax)]; 
de =$[foly tax) +fi(y—a2)] de + [fly tan) +ji(y—an)}au 


_ Joly + ax) (dy +adx) —fi(y — ax) (dy — adz) 
2a 
Hence, z= F(yt+axc)+F,(y— az), 


where F and F, denote arbitrary functions obtained by integrat- 
ing f, and f2, which are arbitrary. 


308 INTEGRAL CALCULUS. [ART. 225. 


225. When a differential equation does not come under any 
of the forms given in the key, a change of dependent or inde- 
pendent variable, or of both, will often reduce it to one of the 
standard forms. No general rule can be laid down for such a 
substitution. It will, however, often suffice to introduce a new 
letter for the sum, or the difference, or the product, or the 
quotient of the variables, or for a power of one or of both. 
Sometimes an ingenious trigonometric substitution is effective, 
or a change from rectangular to polar codrdinates ; that is, the 
introduction of cos @ for « and r sin ¢ for y. 

The following examples of such substitutions are instructive. 


(A.) Change of dependent variable. 





(1) (@+ y= a°. reduces to = @ ys —a?dz=0, 


if we introduce z= @ + y. 





(2) “ = sin(¢ — @), reduces to maa do=0, 
ld 1—sin w 
if w= o>—6. 
(3) (~— y*)dx +2 xydy =0, reduces to (a —z) dx +adz=0, 
igen 
(4) ol sare tt ao — y? = 0, reduces to -+da=0, 
ifz— 4. 
x 
(5) Th ny, reduces to © —nz2=0, 
on 


ifzizs aa 
(B.) Change of independent variable. 
2 
qd) a—- Ge +y=0, reduces to 


dy 


AD =, if = sings 


cos? ZY 78 Y 4 sing cos 0 


Cuap. XVIIl.] DIFFERENTIAL EQUATIONS. KEY. 309 


2 2 
(2 ped + fanee! + cos*z.y = 0, reduces to ay +y=0, 
dx dx dx? 
if 2= sin x. 


(C.) Change of both variables. 





(1) (1- te)? ay = YY (2 y’ — a’), reduces to 
ar dx 
v—Z—- — — 2— — a2) =0 » lf 2-07 and y= 7". 
+ dy 3 
(2) (y—«a) (1+ 2°)” hes (1+ y°)~, reduces to 
x 
sin (¢? —0)d¢=d6, if x= tané and y= tand. 


(38) oe aa y) = = a(1 +9 + 4 aA (a? + y?)? , reduces to 


ear Lan lapd if ~=rcos¢d and y=7rsing. 
Vr(1—ar) Va 


310 INTEGRAL CALCULUS. 


KEY. 
— 

Page 

Single equation . . : 82) 0 ga Ts3h0 
System of simultaneous spoauctae Mr bP ek 

I. Involving ordinary derivatives. . ... . II. 310 
Involving partial derivatives .+. . . . . JX ors 

II. Containing two variables . . . rm #8 Bees 


Containing three variables and of fst degree 

General form, Pde+ Qdy+ Rdz=0. . . (86) 327 
Containing more than three cen and of 

the first degree. General form, Pda, + Qda, 

+ Rdag+-- = 0 5 6. 2 0, 8. 


Li, Of first order)! 0. ee gee TV. 310 
Not of first order .°% %) 2.4. |.) o, = 


IV. Of first degree. General form, Moe Ne 0 V. 310 
Not of first degree. . |. .. .> 0) 


V. Of first degree. General form, Mdaz + Ndy 
al Uf 

Variables separated or separable; that is, of 

or reducible to the form Xda + Xdy=0, 

where X is a function of x alone, and Yisa 


function of y alone* .. . lai 
M and N homogeneous Fanieitone nee a ae y of 
the same degree. . ... < .| 2. [Se 


* Of course, X and Y may be constants. 


VI. 


KEY. 


Of the form (ax + by+c) dx + (a'w + b'y+c') dy 


Linear. General form, “ + X,y = X,, where 
a 
X, and X, are functions of x alone* . 
Of the form + X,y= X,y", where X, and X, 
da 


are functions of x alone*. aa rl 
Mdx + Ndy an exact differential. Test, D,M 


Mx — Ny oo" gS PRY A ge RG a UN i 
Of the form F, (wy) ydx + F, (xy) ady = 0 
D,M— D,N 


a a function of x alone 


D,N — D,M 


Saag a function of y alone 


eee DN a function of (ay) 
Ny — Mx 
A solution in the form of a series can always be 


obtained . 


Not of first degree. 
Can be solved as an algebraic equation in ps, 


where p stands for dy : 
dx 


Involves only one of the variables and p, 


where p stands for oy : 
da 


Of the first degree in x and y; that is, of the 
form zfpt+yhp=f,p, where p stands for 
dy 
da 

Of the first degree in x or ; 

Homogeneous relatively to a and ¥ 


* Of course, X, and X, may be constants. 


(3) 


(14) 


(15) 


(16) 


(17) 
(18) 


311 
Page 


314 


314 


316 


317 


317 


317 


318 
318 


312 


VETS 


INTEGRAL CALCULUS. 


Of the form F'(¢, ~¥)=0, where ¢@ and w are 
functions of x, y, and ae, such that d=a 
dx 


w=, will lead, on differentiation, to the 
same differential equation of the second 
order . be Wane 

A single solution will answer . 


Not of first order. 
Linear, with constant coefficients; second 
member zero* . Pered) ©,\! 
Linear, with aan aodeinioutal second 
member not zero * : 
4a "y 


dx n—l 





Of the form (a+ bent ie A(a+ba)” 1 


+...+ Ty= X, where X is a function of 
x alone 


Linear; of second order ; eoaniene not con- 
stant. General form, vt + Pa + Rky=R; 
Q 


P, Q, and # being functions - x 
Either of the primitive variables wanting . 


Of the form i X, X being a function of 
a” 
x alone ft. 


Of the form <= Y, Y being a function of 
dx 











y alone ft. eit 
n ae 
Of the form d ee ie d 
dx an da” y 
Of the form a" oF 3) fi le e/ 3 
da* au? 


Homogeneous on the supposition that a and 


(19) 
(20) 


(21) 
(22) 


(23) 


(24) 
(25) 


(26) 


(27) 
(28) 


(29) 


Page. 


318 
318 


319 


319 


321 


521 
323 


323 


324 
324 


324 


* The first member is supposed to contain only those terms involving the dependent 
variable or its derivatives. 
t See note, p. 310. 


VIIl. 


| >. 


XI. 


KEY. 


y are of the degree 1, te of the degree 0, 
x 


as ae of the degree — 1, 


Bede sous on the supposition that a is of 


the degree 1, y of the degree n, a of the 
8 


AX 
ay 
degree n —1, a2 of the degree n — 2, 
x 
ald ay he 


de’ da?’ 
Containing the first power only of the deriva- 


tive ot the highest order 


Of the form — a Y mice xo as | |= 0, where 


Homogeneous relatively to y, 





de dx 
X is a function of # alone and Y a func- 
tion of yalone* . . » 
Singular integral will answer 


Simultaneous equations of the first order . 
Not of the first order 


All the partial derivatives taken with respect 
to one of the independent variables . 

Of the first order and, Linear 

Of the first order and not Linear ; 

Of the second order and containing the au 
atives of the second order only in the first 
degree. General form RD2 + SD,D,z+ 
TD/7z=V, where R, S, T, and V may be 
functions of 2, y, z, D,z, and D,z ; 


. Containing three variables 


Containing more than three variables . 


Containing three variables : 
Containing more than three variables . 


* See note, p. 310, 


(30) 


815 


Page 


331 


330 
330 


330 
ddl 


314 


(1) 


(2) 


(3) 


(4) 


INTEGRAL CALCULUS. 


Of or reducible to the form Xda-+ Ydy = 0, where X is 
a function of 2 alone and Y is a function of y alone. 

Integrate each term separately, and write the sum of 
their integrals equai to an arbitrary constant. 


M and N homogeneous functions of » and y of the 
same degree. | 

Introduce in place of y the new variable v defined by 
the equation y= vz, and the equation thus obtained can 
be solved by (1). 


Or, multiply the equation through by ae and its 
ss y 


first member will become an exact differential, and the 
solution may be obtained by (6). 


Of the form (aa + by +c) dx+(a'e+bd'y +c’) dy=0. 


If ab'—a'b=0, the equation may be thrown into the 
form (ax + by +c) daw + gs (ax+by+c)dy=0. If now 
a 


z= ax-+ by be introduced in place of either 2 or y, the 
resulting equation can be solved by (1). 

If ab'—a'b does not equal zero, the equation can be 
made homogeneous by assuming « = #'—a, y= y'—8, and 
determining a and £ so that the constant terms in the new 
values of M and WN shall disappear, and it can then be 
solved by (2). 


Linear. General form “ + X,y= X,, where X, and 
dx 


X, are functions of x alone. | 

Solve on the supposition that X,=0 by (1); and from 
this solution obtain a value for y, involving of course an 
arbitrary constant C. Substitute this value of y in the 
given equation, regarding CO as a variable, and there will 
result a differential equation, involving C and a, whose 
solution by (1) will express C as a function of w Sub- 
stitute this value for C in the expression already obtained 
for y, and the result will be the required solution. 


(5) 


(6) 


@ 


(8) 


(9) 


(10) 


KEY. 315 


Of the form oh X,y = X,y”", where X, and X, are 
functions of x alone. 

Divide through by y”, and then introduce z= y'” in 
place of y, and the equation will become linear and may 
be solved by (4). 


Mdx + Ndy an exact differential. Test D,M=D,N. 


Find { Maz, regarding zy as constant, and add an arbi- 


trary function of y. Determine this function of y by the 
fact that the differential of the result just mentioned, taken 
on the supposition that # is constant, must equal Ndy. 


Write equal to an arbitrary constant the f Mdx above 


mentioned plus the function of y just determined. 


Maz + Ny = 0. 
Divide the first term of Mdx+ Ndy=0 by Ma, and 
the second by its equal —Vy, and integrate by (1). 


Mx — Ny = 0. 
Divide the first term of Mdx+ Ndy=0 by Ma, and 
the second by its equal Ny, and integrate by (1). 


Of the form f, (wy) ydx +f, (ay) xdy = 0. 

Multiply through by ae and the first member 
will become an exact differential. The solution may then 
be found by (6). 


D,M— D,N 
’ 

N (es DN 
Multiply the equation through by e N , and 
the first member will become an exact differential. The . 
solution may then be found by (6). 





a function of x alone. 





316 


(11) 


(13) 


INTEGRAL CALCULUS. 


D,N — D,M . 
—____¥", a function of y alone. 

M Ape N-DyM , 
Multiply the equation through by e’ ” °™, and 


the first member will become an exact differential. The 
solution may then be found by (6). 
Dy Mira dte Me a function of (ay). 
Ny — Mx Dy M—DzN 
Multiply the equation through by ev Ny—Mz where 
v= ay, and the first member will become an exact differ- 
ential. The solution may thus be found by (6). 


-adv 


A solution of Mdz + Ndy =0 in the form of a series 
can always be obtained. 


Throw the given equation into the form “~=— 
| dx N 


then differentiate, and in the result replace a by 
a 


-=, thus obtaining a value of aig in terms of .& 


and y; by successive differentiations and substitutions 
ay dty 


da®’” dat’ 








get values of etc., in terms of x and y. 


If y is the value of y corresponding to any chosen 
value a of w, y can now be developed by Taylor’s 
Theorem. 

We have y = fe =f (a + 2% — Xp) 


= fy + (@ — %) ft + CA) cs WN hide iy oe a py ae 
or 


y= Tas eral 





is oy (a — a)? a a Yo LE owe: a)* d? Yo 
21 dx, oa dx? 


e Tyo LY 


etc. 
’ b) ° 9 
da, da?’ da?’ 


where 





are obtained by replacing # and y by a and y im the 


values of dy @y @y 


9 09 ’ 
de da dz? 





described above. 


(14) 


(15) 


(16) 


KEY. OTT, 


In the general case wy is entirely arbitrary, and if the 
given equation is at all complicated, the solution is apt to 
be too complicated to be of much service. If, however, 
in a special problem the value of y corresponding to some 
value of x is given, and these values are taken as y% and 
%, the solution will generally be useful. 


Can be solved as an algebraic equation in p, where p 

stands for oy 
dx 

Solve as an algebraic equation in p, and, after trans- 
posing all the terms to the first member, express the first 
member as the product of factors of the first order and 
degree. Write each of these factors separately equal to 
zero, and find its solution in the form V—c=0 by (V.). 
Write the product of the first members of these solutions 
equal to zero, using the same arbitrary constant in each. 


Involves only one of the variables and p, where p stands 


for dy 
da 


By algebraic solution express the variable as an expli- 
cit function of p, and then differentiate through relatively 
to the other variable, regarding p as a new variable and 
remembering that CaaS A There will result a differen- 

Pp 
tial equation of the first order and degree between the 
second variable and p which can be solved by (1). 
Eliminate p between this solution and the given equation, 


and the resulting equation will be the required solution. 


Of the form afp+ yfip =fsp, where p stands for S ‘ 
a 

Differentiate the equation relatively to one of the vari- 

ables, regarding p as a new variable, and, with the aid of 

the given equation, eliminate the other original variable. 


There will result a linear differential equation of the first 


(18) 


(19) 


(20) 


INTEGRAL CALCULUS. 


order between p and the remaining variable, which may be 
simplified by striking out any factor not containing or 

a 
e, and can be solved by (4). Eliminate p between this 
C 


solution and the given equation, and the result will be the 
required solution. 


Of the first degree in x or y. 


The equation can sometimes be solved by the method of 
(16), differentiating relatively to the variable which does | 
not enter to the first degree. 


Homogeneous relatively to x and y. 
Let y= vz, and solve algebraically relatively to p or »v, 
p standing for at. The result will be of the form p= fv, 
a) Wig Ueno a ba | 
p=fo, Wa po, WH py, 22 po= py, 
an equation that can be solved by (1). If 
v= Fp, = Fp, y=aFp, 
an equation that ait be solved by (16). 
Of the form /’(¢, /) = 0, where ¢ and y are functions 
of x, y, and ze such that ¢=a and y=6 will lead, on 


differentiation, to the same differential equations of the 
second order. 


Eliminate a between ¢=a and Y= b, where a and 0 
dx 


are arbitrary constants subject to the relation that 
F(a, b) = 0, and the result will be the required solution. 


Singular solution will answer. 
Let a =, and express p as an explicit function of 2 
aX 


and y. Take GD regarding « as constant, and see 


(21) 


(22) 


KEY. 319 


whether it can be made infinite by writing equal to zero 
any expression involving y. If so, and if the equation 
thus formed will satisfy the given differential equation, it 
is a singular solution. 


dl ( ) 
Or take sl regarding y as constant, and see whether 
dx 


it can be made infinite by writing equal to zero any ex- 
pression involving x. If so, and if the equation thus 
formed is consistent with the given equation, it is a 
singular solution. 


Linear, with constant coefficients. Second member 
zero. 

Assume y =e”; m being constant, substitute in the 
given equation, and then divide through by e”*. There 
will result an algebraic equation in m. Solve this equa- 
tion,‘and the complete value of y will consist of a series 
of terms characterized as follows: For every distinct 
real value of m there will be a term Ce”*; for each pair 
of imaginary values, a+bV—1, a—bV—1, a term 
Ae“ cosba + Be“ sinba; each of the coefticients A, B, and 
C being an arbitrary constant, if the root or pair of roots 
occurs but once; and an algebraic polynomial in # of the 
(r—1)st degree with arbitrary constant coefficients, if 
the root or pair of roots occurs 7 times. 


Linear, with constant coeflicients. Second member not 
zero. 


(a) If a particular solution of the given equation can 
be obtained by inspection, this value plus the value of y 
obtained by (21) on the hypothesis that the second mem- 
ber is zero, will be the complete value of the dependent 
variable. 


320 


INTEGRAL CALCULUS. 


(6) If the second member of the given equation can 
be got rid of by differentiation, or by differentiation and 
elimination between the given and the derived equations, 
solve the new differential equation thus obtained, by (21), 
and determine the superfluous arbitrary constants so that 
the given equation shall be satisfied. 

In determining these superfluous constants, it will 
generally save labor to solve the original equation on 
the hypothesis that its second member is zero, and then 
to strike out from the preceding solution the terms which 
are duplicates of the ones in the second solution before 
proceeding to differentiate, as from the nature of the case 
they would drop out in the course of the work. 


(c) If the given equation is ‘of the second order, solve 
on the hypothesis that the second member is zero, 
by (21), obtain from this solution a simple particular 
solution by letting one of the arbitrary constants equal 
zero and the other equal unity, and let y= v be this last 
solution ; then substitute vz for y in the given equation ; 
there will result a differential equation of the second order 
between x and z in which the dependent variable z will be 


“wanting, and which can be completely solved by (25). 


Substitute the value of z thus obtained in y=vz and 
there will result the required solution of the given equa- 
tion. 


(d) Solve, on the hypothesis that the second member 
is zero, and obtain the complete value of y by (21). 
Denoting the order of the given equation by n, form the 
gy OY ae Then 
ax. das") ae 
differentiate y and each of the values just obtained, re- 
garding the arbitrary constants as new variables, and 
substitute the resulting values in the given equation; and 
by its aid, and that of the n—1 equations of condition 
formed by writing each of the derivatives of the second set, 





m— 1 successive derivatives 


(23) 


(24) 


KEY. ook 
except the nth, equal to the derivative of the same order in 


the first set, determine the arbitrary coefficients and sub- 
stitute their values in the original expression for 7. 


Of the form 





(a + boy Y Ye Aatinya® eS -+ Ly = X, 


dan} 


where X is a function of z alone. 

Assume a+ bx =e’, and change the independent vari- 
able in the given equation so as to introduce ¢ in place of 
«x. The solution can then be obtained by (22). 


Linear; of second order; coefficients not constants. 


General form a+ a a Qy = 


(a) Ifa particular solution y= v of the equation 


can be found by inspection or other means, substitute 
y= vz in the given equation, which will then reduce to 


the form 
d?z dz 
as cL ag 2 , 
of +(25 ik a = R 


and can be solved by (25). Substitute the value of z 
thus found in y=vz, and the result will be the general 
solution of the given equation. 


(6) The substitution of y= vz in the given equation, 
where v is given by the auxiliary differential equation 


pee Maasai ity 
da 


322 


INTEGRAL CALCULUS. 


and can be found by (1), and should be used in the 
simplest possible form, will lead to a differential equation 
in z of the form 


—-+/iz= fh, 
eS =f 


which is often simpler than the original equation. 


(c) The introduction of z in place of the independent 
variable x, 2 being a solution of the auxiliary differential 
equation 

es pee 
dx” dx 





the simpler the better, will reduce the given equation to 
the form 
is e 








which is often simpler than the original equation. 


(d) If the first member of the given equation regarded 
as an operation performed on y can be resolved into the 
product of two operations, the equation can always be 
solved. The conditions of such a resolution are the 
following: let the given equation be 

oy 4 ot wy = Te 


where u, v, w, and RF are functions of «2; this can be 
resolved into 
d d 
Pads «(oe honey) eee 
(5,+4) (: Faas ‘) ’ 
where p, 5 15 and s are functions of a, if 
pr=u, ar+p dr +s)=v, and 93 pn ane 
dla dx 


and the values of p, g, 7, and s can usually be obtained 


(26) 


KEY. 323 


by inspection. We have a to solve pe +qz=fR 
by (4), and then to solve r+ sy m2 DY (4). 


(e) A particular solution of the equation 
ay. dy 
—~+ P“~+ Qy=0 
da? da ila 


can often be obtained by assuming that y is of the form 
sa,,0", m being an integer, substituting this value for y 
in the given equation, writing the sum of the coefficients 
of 2” equal to zero, since the cquation must be identically 
true, and thus obtaining a relation between successive 
coefficients of the assumed series. The simplest set of 
values consistent with this relation should be substituted 
in the assumed value of y, which will then be a particular 
solution of the equation. If this solution can be ex- 
pressed in finite form, the complete solution of the given 
equation can be obtained from it by the method described 
in (24) (a). If, however, two different particular solu- 
tions can be found by the method just described, each 
of them should be multiplied by an arbitrary constant, and 
the sum of these products will be the complete solution 
of the given equation. 


Either of the primitive variables wanting. 

Assume z equal to the derivative of lowest order in the 
equation, and express the equation in terms of z and its 
derivatives with respect to the primitive variable actually 
present, and the order of the resulting equation will be 
lower than that of the given one. 


Of the form ad =X. X being a function of 2 alone. 
a” 


Solve by integrating n times successively with regard 
to x. 
Or solve by (22). 


324 


(27) 


(28) 


(29) 


INTEGRAL CALCULUS. 





Of the form Cy aay being a function of y alone. 
da? 
Multiply by 9 ty and integrate relatively to x. There 
wv 2 
will result the equation Ge =? yf Ydy + C, whence 
a 
ct (2 f Ydy + C)?, an equation that may be solved 
as 
by (1). 
n n—l 
Of the form Bay as J. 
da” Cae 


Assume 
n—-1 


Oe Y—s% then de % or de al? re CG. 
Nh ros ena dx : ES iB 


After effecting this integration, express 2 in terms of x 








n—1 n—1 
and C. Then, since ee Z, f '— F(x, C), an 
Ota haar 
equation that may be treated by (26). 
Or, since 
ae ay ayity {2 dz 
——“= 2%, —_“_= | zdv+c= | —-+¢, since dv =—. 
da=1 —” dan? We fz Hoh Sz 


Ghee zdz ~— (ee os 
ete al Be aA a te) tem 


Continue this process until y is expressed in terms of 
z and n — 1, arbitrary constants, and then eliminate z by 
the aid of the equation «= iff Z + C. 

z 


d"y aay 
Of the f i E 
e form Bae pas 








Let 





n—2 
“ 2 =z, and the equation becomes a= jz, and 
tn 


may be solved by (27). 


(30) 


(31) 


(32) 


(33) 


KEY. 325 


Homogeneous on the supposition that x and y are of the 
d Ys Ss 
degree 1, oo of the degree 0, —f of the degree — 1, 


Assume «= e®, y=e%z, and by changing the variables 
introduce @ and z into the equation in the place of # and y. 
Divide through by e’ and there will result an equation 
dzaa*2 
involving only z, —, —,---, whose order may be de- 
6 5 ie de de? ° y 
pressed by (25). 


Homogeneous on the supposition that a is of the degree 


d’y of the 


2 
WA 





1, y of the degree n, i of the degree n —1, 
da 


degree n — 2,---. 

Assume «= e°, y=e"®z, and by changing the variables 
introduce @\and z into the equation in the place of # and y. 
The resulting equation may be freed from @ by division 
and treated by (25). 


dy ay... 
da’ dae’ 


Assume y =e*, and substitute in the given equation. 


Homogeneous relatively to y, 


- Divide through by e’ and treat by (25). 


Containing the first power only of the derivative of the 
highest order. 

The equation may be exact. 

Call its first member = If nis the order of the equation, 
Y 





represent oe by p and < - 2, by = Multiply the term 


containing 2 by da and integrate it as if p were the only 


ory 


variable, Paice the result U,; then replacing p by aan? 


326 


(34) 


(35) 


INTEGRAL CALCULUS. 


find the complete derivative Be ees and form the expression 
ae NREL representing it by clay If as contains the 
de: “ida 1 le dx 


first power only of the highest derivative of y, it may 
itself be an exact derivative, and is to be treated pre- 


cisely as the first member of the given equation gee has 
j My 


d Vea 
da 


been. Continue this process until a remainder of 


the first order occurs. 

Write this equal to zero, and see if the equation thus 
formed is exact, see (6). If so, solve it by (Ge 
throwing its solution into the form V, 4a~=@. JA 
complete first integral of the given equation will be 
U,+ U,+-:-+V,1=C.’ The occurrence at any step 


k 


of the process of a remainder td containing a higher 
ax 


power than the first of its highest derivative of y, or the 
failure of the resulting equation of the first order above 
described to be exact, shows that the first member of the 
given equation was not an exact derivative, and that this 
method will not apply. 





xy | dy =e where X is a 
dx da 
function of # fie and Y a function of y alone. Multiply 


1 
through by a and the equation will become exact, 
x 


and may be solved by (33). 


Wad integral will answer. 





Call @ a i) p, and ts g, and find a regarding p and q 


=l 
as the only variables, and see whether - can be made 


infinite by writing equal to zero any factor containing p. 


(36) 


KEY. BT 


If so, eliminate q between this equation and the given 
equation, and if the result is a solution it will be a singular 
integral. 


General form, Pda + Qdy + Rdz=0. 

If the equation can be reduced to the form Xdx + Ydy 
+ Zdz=0, where X is a function of # alone, Y a function 
of y alone, and Z a function of z alone, integrate each 
term separately, and write the sum of the integrals equal 
to an arbitrary constant. 

If not, integrate the equation by (V.) on the supposition 
that one of the variables is constant and its differential 
zero, writing an arbitrary function of that variable in place 
of the arbitrary constant in the result. Transpose all the 
terms to the first member, and then take its complete 
differential, regarding all the original variables as variable, 
and write it equal to the first member of the given equa- 
tion, and from this equation of condition determine the 
arbitrary function. Substitute for the arbitrary function 
in the first integral its value thus determined, and the 
result will be the solution required. 

If the equation of condition contains any other varia- 
bles than the one involved in the arbitrary function, they 
must be eliminated by the aid of the primitive equation 
already obtained; and if this elimination cannot be per- 
formed, the given equation is not derivable from a single 
primitive equation, but must have come from two simul- 
taneous primitive equations. 

In that case, assume any arbitrary equation f(2,y,z) =0 
as one primitive, differentiate it, and eliminate between it 
its derived equation and the given equation, one variable, 
and its differential. There will result a differential equa- 
tion containing only two variables, which may be solved 
by (III.), and will lead to the second primitive of the 
given equation. 


328 


(37) 


( 


8) 


INTEGRAL CALCULUS. 


General form, Pda, + Qda, + Rda, + + = 

If the equation can be reduced to the form X,da,+ Xda, 
+ Xl + +++ = 0, where X, is a function of x, alone, X, 
a function of a, alone, X, a function of x, alone, etc., inte- 
grate each term separately, and write the sum of their 
integrals equal to an arbitrary constant. 

If not, integrate the equation by (V.), on the supposi- 
tion that all the variables but two are constant and their 
differentials zero, writing an arbitrary function of these 
variables in place of the arbitrary constant in the result. 
Transpose all the terms to the first member, and then 
take its complete differential, regarding all of the original 
variables as variable, and write it equal to the first mem- 
ber of the given equation, and from this equation of con- 
dition determine the arbitrary function. Substitute for 
the arbitrary function in the first integral its value thus 
determined, and the result will be the solution required. 

If the equation of condition cannot, even with the aid 
of the primitive equation first obtained, be thrown into a 
form where the complete differential of the arbitrary func- 
tion 1s given equal to an exact differential, the function 
cannot be determined, and the given equation is not deriv- 
able from a single primitive equation. 


System of simultaneous equations of the first order. 

If any of the equations of the set can be integrated 
separately by (II.) so as to lead to single primitives, the 
problem can be simplified; for by the aid of these primi- 
tives a number of variables equal to the number of solved 
equations can be eliminated from the remaining equations 
of the series, and there will be formed a simpler set of 
simultaneous equations whose primitives, together with the 
prinitives already found, will form the primitive system 
of the given equations. 

There must be » equations connecting » +1 variables, 
in order that the system may be determinate. 

Let 2, 2, a, ....., %, be the original variables. Choose 


(39) 


(40) 


KEY. 329 


any two, x and 2, as the independent and the principal de- 
pendent variable, and by successive eliminations form the 


da, ng ; 
Ades » up to oem Sn (@5X,%o, .....,%,). Differentiate the first 
dle 
of these with respect to « n—1 times, substituting for 
di, dx dx : ; 
—*, —3,.....,—", after each step their values in terms of 
dx dx da 
the original variables. There will result n equations, 
which will express each of the successive derivatives 
2 nn 3p 
yea a, ao; oe; 
i eR 
alr” 





eel COPING Ol Ua Dyy, Was) te:46 Pa 








da’ dx?” da?’ 
Eliminate from these all the variables except aw and a, 
obtaining a single equation of the nth order between « 
and x. Solve this by (VII.), and so get a value of a, in 
terms of x and x arbitrary constants. Find by differen- 
aay Tey ve : at and write 
ei kon ant 
them equal to the ones already obtained for them in terms 
of the original variables. The » —1 equations thus formed, 
together with the equation expressing 2, in terms of x and 
arbitrary constants, are the complete primitive system 
required. 


tiating this result values for 





System of simultaneous equations not of the first order. 

Regard each derivative of each dependent variable, 
from the first to the next to the highest as a new variable, 
and the given equations, together with the equations de- 
fining these new variables, will form a system of simulta- 
neous equations of the first order which may be solved by 
(38). Eliminate the new variables representing the 
various derivatives from the equations of the solution, and 
the equations obtained will be the complete primitive sys- 
tem required. 


All the partial derivatives taken with respect to one of 
the independent variables. 


330 


(41) 


(42) 


(43) 


INTEGRAL CALCULUS. 


Integrate by (II.) as if that one were the only indepen- 
dent variable, replacing each arbitrary constant by an 
arbitrary function of the other independent variables. 


Of the first order and linear, containing three variables. 
General form, PD,z + QD,z = f. 


Form the auxiliary system of ordinary differential equa- 


tions 2% — in =. and integrate by (38). Express their 


primitives in the form w=a, v=b, a and b being arbi- 
trary constants ; and vu = fv, where fis an arbitrary func- 
tion, will be the required solution. 


Of the first order and linear, containing more than three 
variables. General form, P,Dz,2 + P,Dz,% +--+ = R, 
where 2%, %, , %, are the independent and z the depen- 
dent variables. 

Form the auxiliary system of ordinary differential equa- 


Cay _ ity _ Ay __ oe and integrate them by (38). 
Be ee Pah 


Express their primitives in the form v, = a, %.= 0b, v;=¢, 
nese , and 0, = f(V2,V35 ****75U,), Where f is an arbitrary func- 
tion, will be the required solution. 


tions —— 





Of the first order and not linear, containing three varia- 
bles, F'(x,y,2,p,q¢) =0, where p= D,2z, q = D,z. 

Express g in terms of a, 7,2 and p from the given equa- 
tion, and substitute its value thus obtained in the auxil- 


iary system of ordinary differential equations <= dy 
ei er CAD dp 

q—pD,q D,q+pD.4 
these equations, by (36), a value of p involving an arbi- 
trary constant, and substitute it with the corresponding 
value of ¢ in the equation dz = pdx + qdy. Integrate 
this result by (36), if possible; and if a single primitive 
equation be obtained, it will be a complete primitive of the 
given equation. 


Pp 
Deduce by integration from 


KEY. 331 


A singular solution may be obtained by finding the 
partial derivatives D,z and D,z from the given equation, 
writing them separately equal to zero, and eliminating p 
and q between them and the given equation. 


(44) Of the first order and not linear, containing more than 
three variables. F'(x1,%2, +++++, Uy 25 P15 Poy °°") Pn) = O, where 
Pi= Dz,2, Po= Dz,2, 0° : 

Form the linear partial differential equation 3;[ (Dz; F 
+ pi Dz F) Dp & — Dp, F(Dz,2 + piDz®) | = 0, where ® is 
an unknown function of (2, -*:- > Xn Pris *****7 Pn), and where 
>; means the sum of all the terms of the given form that 
can be obtained by giving 7 successively the values 1, 2, 
Bae in. 

Form, by (42), its auxiliary system of ordinary differen- 
tial equations, and from them get, by (38), n —1 inte- 
corals, B, = ay, By = Ay, -*---,®,_; = A,_}- By these equations 
and the given equation express )), Ps, **- > Pn in terms of 
the original variables, and substitute their values in the 
equation dz = p, da, + p,da, + -* +p,dx,. Integrate this 
by (37), and the result will be the required complete primi- 
tive. 


(45) Of the second order and containing the derivatives of 
the second order only in the first degree. General form, 
RDZ2z+SD,D,z + TD72z=V, where R, S, T, and V may 
be functions of x, y, z, D,z, and D,z. 

Call D,z p and D,z q. 
Form first the equation 
Rady’ — Sdxdy + Tdx’? = 0, [1] 


and resolve it, supposing the first member not a complete 
square, into two equations of the form 


dy —m,dx= 0, dy —m,dx = 0. [2] 
From the first of these, and from the equation 


Rdpdy + Tdqdx — Vdxdy = 0, [3] 


INTEGRAL CALCULUS. 
combined if needful with the equation 


dz = pdx + qdy, 


seek to obtain two integrals u.=a, v,= £8. Proceed- 
ing in the same way with the second equation of [2], 
seek two other integrals u,=a,, v,= ,; then the first in- 
tegrals of the proposed equation will be 


m= fir, Ue = fo Vo, [4] 


where f, and f; denote arbitrary functions. 

To deduce the final integral, we must either integrate 
one of these, or, determining from the two p and q in terms 
of x, y, and z, substitute those values in the equation 


dz = pdx + qdy, 


which will then become integrable. Its solution will give 
the final integral sought. } 

If the values of m, and mz, are equal, only one first in- 
tegral will be obtained, and the final solution must be 
sought by its integration. 

When it is not possible so to combine the auxiliary 
equations as to obtain two auxiliary integrals w= a, v=8, 
no first integral of the proposed equation exists, and this 
method of solution fails. 


RY cy 333 


EXAMPLES. 


(1) sinvxcosy.dw—cosxsiny.dy=0. Ans. cosy=ccos2. 


(2) (w+ y)rot = at. Ans. (1—a@)e+y—atan12tY — ¢. 


a 

(3) a= 8in (— 8), Ans. otn| 7-25 = ote, 

(4) oY ytave—y=0. Ans. sin? =e —-a. 
dx x 


(6) y—a ate Hoa+yyi. 
Ans. ctn Fi _ : (tanty — tanta) = tan-!y +c. 
dy dy\" | 9 2\ 3 
Brey yal la (Oe 2 
(8) (#Ft—») =al 14 (74) Jat +9 
Ans. 2a (x+y?) = (a +y’)? —xcose+y sine. 
(7) [2/ (ey) — 2] dy + ydx = 0. Ans. y= ce~vi. 


(8) (7w—y)+ 2 ay ot = Ne Ans. wes =c. 


(9) (2e—y+1)dx+(2y—x%—1)dy=0. 
Ans. ww —ay+y+u—y=e. 


(10) a + ycosx%= Ans. y=sinxg —1+ ce", 
a 


sin2 a 
Py 

| (11) (1— at) SY — ay = amy. Ans. y= [eV (1 —2’)—a]1. 
dy 1 mat 
1 ay cai], Ans. —-=2—y7° ; 
(12) ay ( eeu ew cho y+ ce? 


(13) y (att y+ at) Y + 0(a? + yf — a?) =0. 
Ans. (2? + 7°)? —207 (#2 —y*)=c. 


334 INTEGRAL CALCULUS. 


ady — yda _ a? + uP ae 
(14) YOY asa ar Ans. a Bis 


(15) LU Nenana Ans. (y—alogx—c) (y+aloga—c) =0. 
da a 5 


(16) & 2) + 2yctnatt Tay, 
Ans. (y sin?” — ‘) (u cos?= — e) eth 
2 2 
a7) H(i ty)=2ety). 
Ans. (2y—2x’—c) [log (a+ y—1)+2—c]=0. 


3 2 
(18) to — (+ ay +9) + (abyaty?fay?) UY aby—0. 
x dla las 


Ans. Geet \(e+5 —¢) (logy 5 —) =0. 
y 


2 Y¥\/dyV 2yd 2 
19) (7 mea Ne en a ree 
Ot ( 4 ) ts x dx ds a 


, POST ES fel eer 
Ans. (y+ tog oot ee it “) (y ese log te ae pie, :) = 0. 
Yue y 





2 ; 

(20) He SiN iba: dy ° Ans. y=cut+e—c’. 

dey dx \dax 2-1)? 

Singular solution, y= Et 

(21) y= (Ze UY 4 oat Ans. y?=2ca+. 

dx 
‘ dy dy 
22 = (2? — ay "9, 
( ) 3 -(@) f= (x? Ra ae 
2 
Ans. y?— ca? +" = 
: Y t5 ri: 
(23) y= 2att te at): Ans. y2=2ca+c. 


(24) a(n taut + a= 0, Ans. ?+cey+aa=0. 


KEY. ~ 335 


(25) f[a(Gt) |+29dt —y =o. 
ah % “Ans. (b+y)?=4aa, f(a) +b =0. 





2 
(26) #— s =1 9 = vt | Ans. sat a b, f(a) +b=0. 
dx 


dty ,@y,.@y_ ,dy 
24) == —4-—~ 4+ §6—+ —4— = 0. 
te as | oae  a 
Ans. Y = (Co +, % + CoH? + 63.0%) €*. 
{28) WY 9449 4 Wy =e Ans. y=(q+¢ eye © 
dx? dx ; : (k—1) 


dty ad’y 
29) == + 2—+ == 
( eet at ny is 
Ans. y= 7 +(A+ Br) cosx+(C+ Dx) sing. 


30) EY _ 9 nip 
(30) meet Ue COS &. 


Ans. y= Ae*™ + o| (2 — a COs @ +(¢ + a sina | 
dy o@y , @y | 
eg Se OY 
a) dat da? | dae 
4 
Ans. y= (A + Br) e+ (C+ Da) +120? + 808 +5 + 





a? 
20. 
2 
(32) ee ey 4 4 y a. 
dar’ dae Ans. y=(A+ Ba) e* + 4(20° +424 38), 


2 
(33) ot + y= cose. Ans. y= Acosz+ Bsinz +> sine. 
a 


(34) Ut dy = 2 sin? 2. 
ANS. of = (A—g5)eos2e+(B,)sin22 +6 


OU yaa 
(35) Rss ob 2y = eloge. 
_ Ans. y= Ax cos (log) + Basin (logx) + x loge. 


(40) 


(41) 


(42) 


(43) 


(44) 


(45) 


(46) 


INTEGRAL CALCULUS. 


3 
y ety _ etl +208 v4 ae 


da? da? a 1 ) 
Ans. y= «(A+ Blog) + Cx sea — 33 a 


ay _ 2 dy 2 1 3 

baa Ak Wipe = (). Ans. y = —(Ae™ + Be-"). 

pea Witeen cancnee ns. 9 = he 

2 

“4+ tana! + costa. Of mm 

“ Ans. y = Acos (sinx) + Bsin (sina). 
ay 

Le are =a); 

( oy Coat 





Ans: y= —# (A+ Blogs eet 
— 2 


2 
(1 + at) CY 200d 4 2y = 0. 
i is Ans. y= Be+ A(1—2’). 


iy my ae ty ay 1 


; os it" Ls 
de? x—-1ldze. «#—1 


Ans. y= Ae* + Br —(1 +2"). 








oot 2a(L+a) 4 2(14a)y=a%, 
ANS ye Avet + Br 


sin ot — 2y = 0. Ans. y=Actna+ B(1—-2#ctnz2). 
a Tad tab ae. 2 bloga . 
dar” ae NY i 
Ans. y = Aloge + & loge = B(loge ara} 
log # 
ay a dy ug Ne ZS Ane, 
da? (m et (” ayn ee 





+ 


Ans. y=e™| A norton 
Pris o( + iT 9@a—1) 


2 
Pepe mech ACh 
iaies Ans. y=2x(Acosax+Bsinaz). 


B e ) 


KEY. - 337 


2 
(47) TY 2bal + paty= 0. 
" Ans. y=e? (AcosevVb + Bsinaz Vb). 





2 
(48) wt — 4a wil + (4a? — 8) y ae". 
Ans. y= e* (Ae? + Be *—1). 
2 
(49) (1 at) Sh a —(1 Es ese 
Ans. 2 os 1_(@+Acosa+Bsine). 
— x 
(50) eee ey t/a 8 = 0, 


dx” ./y da 4 a? 
Ans. y=e (Ast + =) 


ty zy ee 2netnna ly + (m? — n*?) y =0. 
Ans. y=(A cosme + Bsin mx) escnz. 





(52) (a Fac — cty = 0. 
Ma 1) Bae Vo 1)" 
(53) eae Lhe ek py = 9. Ans. y= Asin® + B cos”. 
df. eda 2* x x 








(54) Py 3se+1 ee ae 6(a+1) |- 


i 2 —1 dz —1)(8a%+5) 
Ans. y=[A+ B log( (a —1)(3a+5))]V(« — 1 (8a+5). 
(55) (1a?) SY ey = 0. 


Ans. y = Ae?tin #4 Be-esin”'s, 
(56) (1 + aa) £4 ae ema Paps 1), 
Ans. y= A(V1 + ax 2 ee 
(57) (@—1) (w—2) © ¥— (22 — 3) B+ 9y=0. 
da dx 
Ans. y=c(x%—2)?+c! (x — 2) [(~ — 2) log (w — 2) —1]. 


338 INTEGRAL CALCULUS. 


(58) (3-2) SY Bice 4a) Os 0. 


Ans. ya ery co(89_ 15 e+ — 7o73) 
2 
(59) (a?— a2) 24 ga% — 19y=0. 
tr dx 2 
Ans. y= oe 
(a—ay' ” (@— ays 
a2 dy ae. 
60 as —_— — = 0 
( Jen aa Ne da (» ayy 
Ans. y — 4 [ A (sinne — nv cosnaz) + B(cosnx + nx sinna) |. 
PY vod 
Ly eee 0, | Ans. y= I, 
(61) set Ae | | ns. y=clogz+ec 


(62) 9a 4 aty TY 44 4.3? “ LV’ 4 20y AU 0. 
da da da 


Ans. + cay=c'a. 
t l 
(63) (# oF ay ae t 2d 2y € ah ay 30ct +y=0. 
“ Find a first integral. 
Ans. ocd +/ y +ay=c. 


(64) etl + a! 44 Qay— 1) Ley =0. 





Find a first integral. 


Ans. ott fat + xy’=c. 








(65) ie + dy bids Ans. (w—a) (y—b) (¢—¢c) =e. 
x—-a yb — 
(66) (y+tz2)dx+dy+dz=0. Ans. e (y+2) =c. 
(67) ede t= OF ee ei 
dt 4 dt 


_it PH i 
Ans. x= ce He, y=(ct+aq)e 2. 


= 


REY 339 





aa dy 
(68) Tat we = 0, aa 
Ans. x= Asinmt+ Beosmt, x«+y=Ct+D. 
(69) D,z= —_. Ans. eu (x ty +2) = dy. 
“e+zZz 
(70) wzD,2+ yzD,z = xy. Ans. eam to(2), 
Peers 2 = 1. Ans. z=an+240. 
a 


(72) #D2Z2+2ayD,D,2+yD7z2=90. Ans..z2= xd (%)+y(2) 


(78) (D,2)?D22 —2D,zD,2 D,D,z + (D,2)? Dz = 0. 
Ans. y= xdz+ yz. 


(74) D,z.D,D,2—D,z.D?2=9. Ans. «= py + vz. 





APPHNDIX. 





Cuar. V.] INTEGRATION. 65 


CHAPTER V. 
INTEGRATION. 


74. We are now able to extend materially our list of formulas 
for direct integration (Art. 55), one of which may be obtained 
from each of the derivative formulas in our last chapter. The 
following set contains the most important of these : — 





D,logz = : gives fe = log. 
a x 
Da’ =a‘ loga Of Jo log a = a’: 
1) aoe 66 ona ae 
D,sinz = cosx a COS@ == Sit T: 
D,cosx = — sinx ‘6 f,(— sinz) = cosa. 
D, log sing = ctnz ‘cf, cine = logsin x. 
D, log cos% = — tanx 6 f,(— tana) = log cosz. 
PD sin = aoa BS jeg ents oe sin7* x 
V/(L— 2) V(1— 2") 
] 
Dtanz'2= BE fee AT 
1+2 J 1+ 2 
D, vers~12 = pearls Se éé ok ak tee = versa. 
/ (2a — 27) Ve —2) 


The second, fifth, and seventh in the second group can be 
written in the more convenient forms, 


Sc 
fam) 


DIFFERENTIAL CALCULUS. [ART. 765. 


a 


fz0* = ; 
loga 





J, Sine = — cosa; 
J,tanx = — log cosa. 
75. When the expression to be integrated does not come under 
any of the forms in the preceding list, it can often be prepared 
for integration by a suitable change of variable, the new variable, 


of course, being a function of the old. This method is called 
integration by substitution, and is based upon a formula easily 


deduced from D,( Fy) =D, Fy. Dy ; 
which gives immediately 


Fy =f.(D,Fy.Dy)- 


Let u=D, Fy, 
then Fy =f, Uy 
and we have f= SAUD) 


or, interchanging # and y, 
fou = Jy(uD,2). [1] 
For example, required (a+ bx)”. 





Let f= Ot Oe, 
and then Japon) =f 8 = eae by [1]; 
: b b 
D,x an 
b 
1 1 gn+1 
h ba)" = = a 
ence S.(a + bx) a dat 5 cieag 


Cuap. V.] INTEGRATION. 67 


Substituting for z its value, we have 


; 1 (a+ bax)" 
b Cece ee Y 
ae b n+1 
EXAMPLE. 
Find ea Ans. log (a + bev). 


76. Iffx represents a function that can be integrated, f(a+ bx) 
can always be integrated ; for, if 


z=a+ be, 
then Dives sf 
b 
and Saf (0 + de) = ff =fyfeD,0 = + fat 
EXAMPLES. 
Find 
(1) fsinaew. 0, pee 1 00s ae. 
a 
(2) /,cosam. Ans. 1 gin an. 
a 
(3) f,tanagw. 
(4) /.ctnae. 
Pee Required | ——-—-. 
(a — 2) 
1 1 1 
vE-@)] 
Let States , 
a 
then OF eth, 


68 DIFFERENTIAL CALCULUS. [ART. 78. 


1 1 1 1 1 1 
ee SS ef 
a a CESS a ee a 
vL-@| 
eases =sin-!z= sin7! =. 
EXAMPLES. 
Find 
Gas Ans. 
oer ns. q ban ae 
(2) renee Ans. vers—*— 
= : 1 
18. Required eta 
Let 2=e+/(2? +a?) ; 
then Z2—u=/(a?+07), 


2—Qeea+e7%=2?+ a?, 


22e2=—= 2? — a}, 





2 2 D/ 2 
2° — O_ z a 
a/ (2? + a?) = 2 —2 = 2 — ——— = ma 
22 2% 
2 2 
Zz a 
Dea 
Ve 


1 
fee aie 


22 gt a 1 wet 
= ecpnys a2 =S2— = logz = log(# + Va? +’). 








EXAMPLE. 


Find Jao Ans. log(a +V2*— a’). 


Cuap. V.] INTEGRATION. 69 


79. When the expression to be integrated can be factored, the 
required integral can often be obtained by the use of a formula 


deduced from D,(uwv) = uD,v + vD,uU, 
which gives w= f,uD,v+f,vD,u 
or S,uD,v = uw — f,vD,u. [1] 


This method is called integrating by parts. 
(a) For example, required /, loge. 
logx can be regarded as the product of log by 1. 


Call loga =u and 1= D,v, 
1 
then tee 
a 
V=2;5 


and we have 


J,logx =f, llogr = f,uD,v = uw — f,vD,u 


= tlogx aa ies vlog a — a. 
© 


EXAMPLE. 
Find /,xlogz. 
Suggestion: Let logy=u and «= D,v. 


Ans. } 2*(loge _ 5) 
2 2 


80. Required f, sin’. 


Let u=sing and D,v=sina, 
then Du = cos”, 
v= — COS2, 


. > 7 way a 2 ye 
J, sin’'a = — sinxcosx +f, cos’a ; 


70 DIFFERENTIAL CALCULUS. (ART. 81. 


but cos’a2 = 1 — sin’, 
SO J,c0os 4 = f,1—/, sin’x = « — f,sin’a 
and J,sin’x = « — sinx cosa — f,sin’a. 


2 /,sin?a = x — sinxcosa. 


J,Sin’w = 4(a@ — sina cos2). 


EXAMPLES. 


(1) Find /,cos’w. Ans. 5 (e + sinxcos”). 


+n2 
(2) /,sinwcosa. . Ans. ae 


81. Very often both methods described above are required in 
the same integration. 

(a) Required f,sin-*x. 

Let sins a = 7, 


then c= siny $ 
D, x2 = cosy, 


J.Siny 2) es cose 


Let u=y and D,v= cosy; 
then Dek. 
v=siny, 
and 


Jy cosy=ysiny—/,siny=ysiny+cosy=xsin-'x+ ./(1—2"). 


Any inverse or anti-function can be integrated by this method 
if the direct function is integrable. 


(0) Thus, Lf a= fy =S,yD, fy = ufy —L,IY 


where Yue fas. 


Cuap. V.] INTEGRATION. 71 


EXAMPLES. 
(1) Find /,cos~*a. Ans. xcos~'x —./(1 — 2’). 
(2) /,tan-'e. Ans. xtan7*xe — slog(1 + «). 


(3) f,vers"*2.  ~ Ans. («—1) vers~*x +./(2% —2”). 


82. Sometimes an algebraic transformation, either alone or in 
combination with the preceding methods, is useful. 





ey 1 
(a) Required Sea : 


ut CASE as pee 
eG Tae ae ea 


and, by Art. 75 (Ex.), 








1 1] 1 r—a 
Moola ay — los (x | 3 
saa rate a) — log(#+a)] oe Scare 


(6) Required /, \ ( . a} 


N(i3) Jae 7 yao * 0S 








re Shes 2) ey = sn 
fo z ae 2 can be readily obtained by substituting y =(1— 2”), 
and is —./(1— 2%) ; 
hence ENG) = sin~'x —./(1— 2’). 


(c) Required /,,/ (a? — x”). 


V@—8) V@=e) J@=a’ 


12 DIFFERENTIAL CALCULUS. [ART. 83. 


Mab ied iad got LR eee a 
and f(a EEC ET, I (@=8) 


eae e 
whence /,,/(a?—2”)=a?sin on =F J@aay’ by Art. 77; 


but Sel (? — 2) = an] (a? see 
by integration by parts, if we let 
u=./(a?— x) and D,v=1. 


Adding our two equations, we have 


2fr/ (a? — a) = wa) (a? — a) + asin ; 
a 
1 aso te 
and Se (@) — x) = 3 (2Ve— a+ asin), 


EXAMPLES. 
Find 
C1) f(a + a’). 
Ans. : [a./ (2? + a?) + @log(a+ Va? + a?)). 
(2) JV (@ — a’). 
Ans. 5 [x./ (a — a?) — e@log(a+ V2? — a?) |. 


Applications. 


83. To find the area of a segment of a ie 
Let the equation of the circle be 


e+y=ia*, 


and let the required segment be cut off by the double ordinates 
through (a%,Yo) and (#,y). Then the required area 


A= 2f,y+C. 


Cnap. V.] INTEGRATION. 13 





From the equation of the circle, 

y=V (ve — 2"), 
hence A=2/,,/(a—2)+0C; 
and therefore, by Art. 82 (c), 


a 
A=2,/(v7—2)+ sin“! : +. 
As the area is measured from the ordinate y) to the ordinate y, _ 


A= 0 when 2 = % ; 


EG, 


therefore 0 = ar/(a?—2,”) + @ sin- > 


C= — a /(’? — ay) — a’sin7'— 


and we have 


x 
A=2#,/(CW—2#)+¢ BUR — &/ (@’—2") — a’sin—! 
If x = 0, and the segment begins with the axis of Y, 
A=2,/(a — 2’) + a sin7“'— 


If, at the same time, «= a, the segment becomes a semicircle, and 


7a? 


ee Vee 
A=a,/(a’?— a’) + sin- : 5 


The area of the whole circle is za?. 


T4 DIFFERENTIAL CALCULUS. (ART. 84. 


EXAMPLES. 
(1) Show that, in the case of an ellipse, 


ge? 
the area of a segment beginning with any ordinate y is 


ai 


idee E af (a? — a?) + a?sin—12 — Xyr/ (a?— 2") — sin". 
a a 
That if the segment begins with the minor axis, 
A= AG a/ (a? — a?) + csi" 
G a 
That the area of the whole ellipse is zab. 


(2) The area of a segment of the hyperbola 


is A=? [2 (0 a8) — clog (2 +V =a) 
= yr/ (a? — a?) + a? log (a+ Vay? — a”) J. 
If 1 =a, and the segment begins at the vertex, 
a : [x/ (2? — a?) — a? log (a +V2? — a?) + a? loga]. 


84. To find the length of any arc of a circle, the coordinates 
of its extremities being (%,%) and (#,y). 


By Art. 52, $= S/[1+ (D.y)*]. 
From the equation of the circle, 


a + 7 = a?, 


Cuar. V.] INTEGRATION. — 15 





we have 24+ 2yD,y = 0, 
Die 
y 
2 
Eda pL 
y? y 
a 1 ale 
$= /,— =)... “+0. TGS.4 fa 
; Ue ay asin = CATS. (2) 
When v= Xp, Saath. 
: X 
hence 0 =asin—! ae es 
C= — asin a 
a 
: & X 
and s=a(sin- a Sis 2), 
a a 


If x = 0, and the arc is measured from the highest point of the 


circle, §= @sin—4— 
If the are is a quadrant, xw=a, 
ma 


o— asin (1)=-> > 


and the whole circumference = 27a. 


85. To find the length of an arc of the parabola y? = 2mx. 
We have 2yD y= 200; 


De 
y 





J/[1 + (D.y)"] =("3 = ; J (m? + 4") ; 


76 DIFFERENTIAL CALCULUS. [ART. 85. 


poo 1 2 2 ee 1 2 2 - 
sap] om +9 Jaa] ive +¥) Dye 





de. by Art. 73 ; 
m 


sa 4 Ninf => —[y Vink Fi+ m'log(y + Vit + F)]+6. 


by Art. 82, Ex. 1. 
If the arc is measured from the vertex, 


s=0 when y=0; 


0 = |. (m?logm) +0, 
2m 


C= — Smlogm, 


_1[yV/(m’ +7’) y +-/(m? + y’) 
oo oa 5 (eee SEW OB torres ereciiars ¥ 


EXAMPLE. 


Find the length of the arc of the curve a’ = 277” included be- 
tween the origin and the point whose abscissa is 15. 
Ans. 19. 


SHORT TABLE OF INTEGRALS. 


COMPILED BY 


B. O. PEIRCE, 


Houuis PROFESSOR OF MATHEMATICS AND NATURAL PHILOSOPHY 
IN HARVARD UNIVERSITY. 


BOSTON, U.S.A. : 
GINN & COMPANY, PUBLISHERS. 
1889. 







The Smee will be grateful | to any pers 


notice t errors in pare formulas to ir 


bear, ‘ { ; «) se 


. ee 


B.0. PEIRCE, = 
° Harvard College, Cai 





I, FUNDAMENTAL FORMS. 


Ls adx = AX. 


2, f af («dx =a (f(x) de. 


wv 


dx 
S$. fe = 1 e 
x og x 


m+1 


4. farae =” » when m is different from —1. 
m+1 
: fea fae 
Jf alog adx =a’. 
dla 
7. f itn 2. 
1+2 


8. if dst = sin-!a. 
RL a 


9. (__@&__ 
avo? —1 





or 


o 








= sec"!a. 





= versin /2. 


10. = 


11. fcosads = sina. 


12. fsinzac= — COS 2. 


+ FUNDAMENTAL FORMS. 
13. fetnzde = log sin. 

14. fran xde = — log cos x. 

15. fran wv secada = seca. 

16, Jsectade == tan. 


17. fesctade = — ctne. 


In the following formulas, u, v, w, and y represent any 
functions of x: 


18. fe +v+w-+etc.)dx= | udx+ | vdx + f ude + ete. 
190. f ude = uv — J rvdu. 


19d. yt dxz= uv — ye ae, 
da da 


20. iO) dee = fo". 
da: 


RATIONAL ALGEBRAIC FUNCTIONS. 5 


II, RATIONAL ALGEBRAIC FUNCTIONS. 





A.— Expressions INVoLvInG (a+ 02). 


The substitution of y or z for w, where y=az=a-+ ba, 
gives 


21. oan (pa, 


22. fe (a+ bx)"dx= =f yy — a)dy. 


1 
23. fe (a + bx)"da = aay (y—a)"dy. 








4, x” dx ics 1 (y — a)"dy. 

(a+bu)™ On ng 

2. f__ _ pee 
a” (a + ba)™ Me ingitd 

Whence 
dx 1 

26. = —] ba). 

a+bae 0b Te eed, 


2%. = — —______- 
Jesters b (a+ bx) 
28, [—<,.. Moai se 
(a + bx)?® 2b(a+ ba)? 


x dx 1 
29. rar tt a ela + ba) |. 


oo 
—) 


; tees jf los at be) + a+ ba 


6 RATIONAL ALGEBRAIC FUNOTIONS. 


x da 1 1 «a 
aT J (aba) ale ate a+ | 


ae. = . [$(a + ba)? — 2a(a + bx) + a log(a + bx) ]. 


x dx 
= b 
Pa een plot a 


da 1, a+ bx 
4. f—S =F Jog Et. 
: x(a + ba) Las eae 














da 1 1 a+ ba 
pyc | (stadathndatan ets TUN ikS 9 ae D5 
os Wiraasey a(a+bx) @ tas 
dx A b a+ ba 
86. Jac =-= hal eM 
a (a + ba) Rant Gi °8 x 


B. — Expressions Invotvine (a + 02"). 


37. f se oo inate 














dx 1 c+2 
, ——, = — log ——-: 
1 Co art De MaDe 
da 1 Bik: 
39. eae \ fa>0, b>0. 
Nee LUGE an & ane a> ot 
40. f do 1 pg VEteVEd 16 ee 
a+ ba = 24/_ab Si eae 
41 x >: dx 
Ge 2a(a + ba’) J a+ ba? 


ae Ua Si eal WS a a el ee eee 
(a+ ba)" 2ma (ace Een 2ma anseete 


43. f x da eee 2 a\. 
Fae bok a DON ean, 


42. 





da 
a 
J athe 2 (oti Hagen 
da 1 a? 
° —__—____. = — log ——___. 
a | eg 2a -a+t+ba? 
fee ae 
erage ba bo bY a+b" 
47 fo met ~" clar 
* J a (a + b2*) ax a + ba® 
48. GORE aS ST aie Se t f , 
(a+ba?)"*!  2mb(a + ba®)™ ~ 2mb/J (a+ ba®)™ 
49. fo =—=? 1 Se 
(a+ bay" a.) a? (a+ ba®)™ etree 
dar ky (k +2)? Pi 2e—k 
° os aaa ame 0g) Batt ee, 
ig a + ba’ sal? ee + V8 tan kvV3 
where bk? =a. 
da I? — kx +- a? W2a—-k 
51. (3 = = ——| $log( ———*— ) + V3 tan-1=—— |, 
a + bx? sal e( (k +2)? )+ aaa 
where bk? = a. 
52. fm ribs gr” : 
x(a+ bx") an ~a+ ba” 
53 {Se {FS ade 
"J (at bart ad (a+ ba)” (a + bay 
ee (es (ee 
° (a-bar)rt (a+ba")? = (a + ba)? 
55 faa Ee antndae 7 Vo" 24 b {fe dae 
e x(a + ba”)?** a (a + bar”)? »n— "(a +- ba")?t} 


RATIONAL ALGEBRAIC FUNCTIONS. 














RATIONAL ALGEBRAIC FUNCTIONS. 




















ea laa td ee xh, oo 
op) @ EL ee ‘xp J 9% ee zp) 83 

ox ub eXOK eX PETE CAG ee 
Se fee oe 19 o> usys b—K—q4me! { =p.) “82 
x US Exo \ Deeper rhs PSS es 
“ep eG se SO ee me US oe eC ee TY ee tae ey fee 


usy} ‘.q—o07 =b pus w9+%qgq+0V=—X WT 


"(wo + 29+”) DNIATOAN] SNOISSANdXY — "OD 


+d)up 
| ep pea (9 + 1) [oui of (u + du+ Ub) a +a (ut + D) ae ee 





| ap a(utQ =f 1) Tutu x fq (u + du St U) aa os +a (XQ + Dae ev 
= xp alg + v) wt f "99 


U au > Ul WwW 
| ap a Q ae 1”) put a { vdu se a(uXQ ae ») gl jes 
[ates (du z U)q 


| ap aug + ig ere ust wf v(w ae u) i +a(uXQ ae Due wl J 


RATIONAL ALGEBRAIC FUNCTIONS. 9 


68 feast da 
‘ Oty qx qJ X 





ede 2a+tbxe b(2n—1) 








64. — iif 
Xt Mest ng x” 

65 fea=2-2 Dane ree da 
xX c 2c? D.¢ 





66. fen ate .n 2a ue. 
cqX 


he js gm n—m+ 1 b a” 1da 
- J yen (2n—m+1)cX” 2n—m+1 cJ XK 


m — 1 a Cx™-*dx 


In—m+1 ¢ Ginn, 





da 1 x? b da 
: — = — log— — — 
Se Caw. 2a °8X 2a pe 








dx b hy fZ 
° = — log— — — 
fy eX 2a "at ax mos a? as 
70 f i Sa 1 vs deed oat SOS da 
foe tt (m —1)aa"-"X” m—1l ad am lx 
—2npmat cy da j 
m — 1 A ts Be Gara 


D. — RaTIONAL FRACTIONS. 


Every proper fraction can be represented by the general 
form : 
F(#) _ ra" + Goe* + Gee PH oe + In 
F(a) a +hyw 3+ k,v™*? + +) +h, 





a, b, c, ete., are the roots of the equation F(«)=0, so 
that 
F(x) = (a% — a)? (@ — bY (%— Clery 


10 RATIONAL ALGEBRAIC FUNCTIONS. 














AAC obi A, Ay es A, 

ree F(a) (a—a)? («—a)ra a (x — aya agai 
B, B, Be 

Sree (e@—b)1 Gobet recy» 
C; Chien C; a C, 

(Gem Oi eatey a (Ocal ti gee 

+ :-- 


Where the numerators of the separate fractions may be 
determined by the equations 


NS Pilea) f B gr ACR) 
he Let O) Pee chee BB 
=f) e— wy =e b)* 
If a, 6, c, etc., are single rate then p = =F cea, 
and 
S(®) _ A oe B AN 
(v)  &€—a L—-bd. L—C 


F 
pend hee) f (0) 
where <A =F (a)’ B= Fal (by’ ete. 


, etc., etc. 


, etc., efc. 











The simpler fractions, into which the original fraction is 
thus divided, may be integrated by means of the following 
formulas : 





71 hdx  _ ¢hd(mv+n) _ h 
* J (ma + ue m(mea+n)' m(1—1)(ma+n) 
hda 
= - ny 
mae+n — ™m °8 a an Be 


If any of the roots of the equation f(#) =0 are imaginary, 
the parts of the integral which arise from conjugate roots 
can be combined together and the integral brought into a 
real form. The following formula, in which 1=/—1, is 
often useful in combining logarithms of conjugate complex 
quantities : \ 


73. log (w + yi) = $ log (a? +y)+7 tan". 


IRRATIONAL ALGEBRAIC FUNCTIONS. bE 


III. IRRATIONAL ALGEBRAIC FUNCTIONS. 





A.— Expressions Invotvine Va + ba. 


The substitution of a new variable of integration, 
y= Va+ bx, gives 
as 2 
74. [Na+ be de = 35 Vat bay 
75. fava bide = — 20a — Stn) V (a+ ba)? 
150? 
2 
ae fa + bade = 2 (8a — 12aba + 15b?2") V(a + b2)* 
105° 
77. (FFB vate of He 
=f @ Jo, + bar 
da 2Vu+ ba 
ee 
Va + bx b 
ode _2@a—b«) ae 7 
79. Lae ae a ig ge vay a+ ba. 
2 
80. vide _ 2(8a* slndAeataetake a”) earn 
Nat be 7 156 
81. B= Flog ERS), for a> 0. 
e/atba va Va + ba + Va 
82. ae ee ey a ae fora< 0. 
tJatbe V—a —a 
ee Hise Ce A OED EID, dx 





ax 2a t/q be 


12 IRRATIONAL ALGEBRAIC FUNCTIONS. 


2in 
2 A 2 (a+ ba) ?- 

; ba) *3 de = “fysrd 
84 fat n)*ide == {ydy = ces 


7 hein] fap s _a(a+bayt } 
eS Jer =) de Al 4+n 2rn 

x” dx _ 2a™Vat ba _ 2ma x” da 
Vatba (2m+1)b (2m+1)bY Va+ bz 


f Me Ne Vatbe  — (2Qn—38)b da 
a” 


86. 


87. 


Vatoa (n—l1)au”" (2n—2)a of), faa 


@ 


n n—2 
a n—--2 
wv 


89. {——. da =i (—_“— De. dx 


x (a + bx)? «(a+ ba) 2 z (oe 


B. — Expressions* Invotvine Va? + @ anp Va? — 2a’. 


90. {vez oda =4[avVart @ + a? log (a+ Va?+ a’)]. 


91. fve= att lee i(2Ve— a? + a? sin) 
a 








dx —~—; 
92. —_——— = log (* + Veet a’). 
Va? + a? 
da ‘ x 
93. Adal ae I SN ils 
Va? — a? a 


94. {—— Bae Loe 

Cr/y?—Q? a a 

* Gil ee 

ne (eee ae (Levers), 

w@/q2 + a? a wv 

CIE OP UN ek ee 
96. eee dz= Vat gn? oleh, logit Vai £ af, 
x x 


*'These equations are all special cases of more general equations given in the next section. 


IRRATIONAL ALGEBRAIC FUNCTIONS. 


PEN Cer 
97. a da =V22— a? — a cos}%. 
x Hy 








oo SG pe rj 
Vat 

99. f a x VANES STAT 
Va? — a2 





100. {av + oF da=4V (x7 + a*)?. 
101. | eVa? — 2? dz = —1V (a? — 2?)3. 


102. f V (a? + a?) dx 


13 


it tt SONS SE Devel ft gee 4 eee Sea 
=| ove ot) fis ws Es a? + = log (@ +Vor+ #) | 


108. { Ve —2")'dx 


=4{ Ve oar)? F ee a” + oa sin? 


M9 + 2@ 
104, ( —__ = ere ee 
Veta)? CVF +a? 
105, ( —“"__ = = 
V(e@—e)t UV/e— 2 
Cait) ak 


a 
Vietita’)® Vata? 


dla 1 
107, {22 _ = 
(at 2t)® Vaz — a2 


108. {eV @ + a’)? da = 4-V (a? + a?) 





109, (2 V(@ = dx = — VF =a, 


14 IRRATIONAL ALGEBRAIC FUNCTIONS. 
110. feet a? da 
rads wh ent 2 pat Sawn a eee 
= 1V@ ae a bas ; (aVa?+ a? + a? log (« + Va? + a?)). 


ii. fovea? de 
=— iV (@—#Y ot) * seve ig? + a sing 
a 





2 aPis. A et pa BS 
12. {§ =i Vee a? + Zlog (a + Va?+ a’). 





lea ah Ff 
Wie dan ead 2 

113. (2 @__ _ * ere sin”. 

Va? — a? 2 2 
114. af! de _ 4 Veta’ 

oP a? + ar ax 
1i5. f= ores ey 

a /q? — a? a? a 


116, (N@sa de | NO EO Tooke VBE @). 


2 2 2 2 

a — @ Va as 5S ves hee 

re — gin-12. 
a 


x 





118: (eee 1 JFia 


iv 
119. [2s C108 yh OS Rie TY eects 
A Cae ar )8 9 fates 


C.— Expressions Invotvine Va + ba + Cat, 


Let X=a+be+ ca’, g=4ac—b’, and k= =. In order 
to rationalize the function f(a, Va+ be + cx®) we may put 
Va + bx + ca* = VtcVA+ Bx + 2, according as ¢ is positive 
or negative, and then substitute for « a new variable z, such 
that 


IRRATIONAL ALGEBRAIC FUNCTIONS. 15 


2=VA+Be+ eo — 2, if c>0. 
~ 8s ife<0 and >0. 








e. where a and £ are the roots of the equation 


Pear Lisa if c< 0 and ly al 
—c 


By rationalization, or by the aid of reduction formulas, may 
be obtained the values of the following integrals : 











dx 1 7 zs b . 
120. Fe = alog( VE +2Ve+ i Ce Oy 
VX Ve 2Ve/’ 
121. Oe = are eee ifc< 0. 
KAO hi C Vb? — 4ac 


de _ 2(2cx+b). 

an x qv X 

123. Be six) 

xen x Boyv XO 

Ma 12 (Qcr4- bye 2k(n —1) da 


122. 








124, eC ER In—1 J xnasxX 
126. I ers = a i +3 VE 
12%. (X*V Kae = C@EDVY (x4 eae ives VE 


is.) ( X--V Xan — Ce +d) XV _Qn41  (X*de 
4(n+1)e Q(n+1)kJ SX 


mene MENS 1B) dan 


16 IRRATIONAL ALGEBRAIC FUNCTIONS. 





130. ade _ 2 (bu+2a). 











XA qVX 
131 f x da nein Al x Sind dx ‘ 
XV Qn=eXt 2) xox 
x" dx ees OF =>  ob°’—4ac ( ae 
132. acly ts Seb hadnt {cv 
ae as os ‘i 8c IX 
2 eee 
138, f x be (2b 4ac)e+2ab 1 ae 
NEN EN cgqVX cs /X 


184 f oo? da CRO es dac)e +2ab , dact+(2n—8)b" de 
h m/X (Qn—1)eqgX*1VxX  (2n—1) cy Xe 








x? da: =e" 5ba , 5b? ae 


— Bab sabe dx 
. oF 28 \ x4 ee 
si JX OO MLC RO CRO, ee ae) /X 
136, [2 VXde=* YX _b FXa0. 
3C¢ 2¢ 





187, (2 XVXae oa =~ 5. [XVXax, 


138. ee da) CXR 0, xX” dan 
ime dG hoe IX 


Ae ; ne 
139. {a VXde = eee XVX , 5b’—4ae A Ba PS 
6c/ 4c 16¢’ 








eX "de wX"VX (2n+38)b (xX*dax 
VK Antle 4(n+1ceJ VX 
fi II (Be 
2(n+1jer VX 


os Toe , 850° 2a\ XVX 
14 {eV Xde = ae en 
, ( 8c + Be S¢y de 


3.ab 
Xa 
Desa aa) es i 





i40. 











IRRATIONAL ALGEBRAIC FUNCTIONS.» 




















17 















































143, f- ee iog| YEN ey TOs 
ar/ X Va x 2a 
dx 1 nf ae ba + 2a f 
143. { —— = ——— sin if pe ay if.a < 0: 
myx) Va aVb? — 4a¢ 
Pe NX ie a 0, 
av xX ba 
145 Bree NX 1 ee dx. 
eX°VE (Qn—1)aX*' a) gyeR 2d X/X 
146. ee EV XO. da 
aa! X ase av X 
47, (EM _ yxy! oe +af. dx 
VX av xX 
xX" dx xX” Da AT NTT Rae eae) fs 
148. tao =O = RP Bs 
av xX oe BENE De Meas Af Xe 
an eS i NE VX 
150 f x” dix =) (= am*de 6b (-a™*da “oS a? dae 
"J xnJ/X Xn1JVX cf} X-/X XV 
ae ge edt XX Reece am X" da 
Beal x (2n+m)c 2c(2n+™m) ee 5 aah 
a — Da Cae Xda 
(2n + m)c ea 
ee 
am Kr X Se Des 
_ (2n4+2m—3)b _ Qn+m—2)e 
2a(m—1) ees al (m—l)a ie 


18 JRRATIONAL ALGEBRAIC FUNCTIONS. 


fg @ Ss lea ae SE Xede 
om VX (m—1) a" 2(m—1)d gm1fX 
m—1 gm 2 1X 








> 


D.— MISCELLANEOUS EXPRESSIONS. 


154, [ V2ax— ade =2—"V2ae—# +S sin 4. 
: a 


f da : x 
155. {= = versin~!“. 
V2ax — 2 a 








157. 7 da peel ae 
(e—1)Ve—1 Ne =1 


158. f+ 2 ae = sina — V1 — 22. 
—2 





159. f 2+ 4ar— V (a + a) (@+ b) 
+(a—b) log (Va+a+Va+5b). 


dx Cor a oa. 
OO leer = sin VE 








161. Re EG! ED Rene B (a+ ba), 
V (a+ bx) (a— Bx) VbB aB + ba 


162. {va + bada = a V (a +ba)*. 


dx Ud Hey acne coe eee 
168, | ————— = * V(a+bz)* 
Vatoe 2b 


IRRATIONAL ALGEBRAIC FUNCTIONS. 





164. { - xa ___ 3(8a—2bx) V (a+ be)?. 
Va + be 10° 
165. (—_. ee seo"#(=). 
a Vx" — a? an ah 
see. (27 1 op VO tM. 


2 AE S aa 
ava ta an Vai tata 


19 


20 TRANSCENDENTAL FUNCTIONS. 


IV, TRANSCENDENTAL FUNCTIONS. 


167. | sinada = — cosa. 
168. | sin’ ax dx = —4cose sina + $a. 
169. f sin’ada = — cosa (sin’x + 2). 
n”- 1 
sin” & COS ® —1 , 
170. fsintcae = — + n—* { sint*ede. 
n n 
ene, sing. 
172. { costede = dsinw cosx+ 4a. 


173. { cos" wda = 4sinx(cos*a + 2). 


fs 1 : n—1l 
174. | cos"*xdxw = — cos"! a sinxz + —— | cos” *ada. 
n n 


175. f sina cosada = 4sin?a. 


176. fsinte cos’ ada = —4(fsindu—2). 
~ m+, 
f _cos™th a 
177. sinwz cos” “da = 
m+ eel scltt 
; sin™ +l 
178. f sin” cos x2dz = ———_-- 
; m+1 
, : cos” -!a sin”t! & 
179. cos” x sin” #d2 = ———__——" 
m+n 
m—l fy! : 
4 ma { cos” 2” sin” x da. 
m+n " 
; sin”!a cos™tla 
180. | cos” x sin" xdz = — ————_— 
m+n 
i —~ 1 


cos” a sin"-? «da. 





m+n 


TRANSCENDENTAL FUNCTIONS. 








181. foes ede | cos™*1a m—n+2 (cos*xdx 
sin"®  (n—1)sin" a n—1 sin*? 7a 

182. freee ada _ cos™—! a 4 m— a dla 
sin"x  (m—n)sin"1z m—n sin” & 


Cos" & sin" (= 


sg, (sin™ade _ SG) iGa 
eo G- 





184. {: ae de 
7 sin” x cos” x 














21 


Xoo See 1 m+tn—2 da 
~~ na—1— sin™ 1x. cos") n—1 J sin™2. cos*-?% 
Byer 1 m++tn—2 dx ; 
~ m—1- sin™a. cos" a m—1 sin™ *a.cos"% 
185. ODgs lhe od) i CORR m — 2 dx 
sin” # m—1 sin™!2 m—1/ sin”™-*a 








186. f dx th 1 _ sing ine n— nae f ‘aca 
cos" z 2—1 .cos**2 xn—1/7 cos* 2 
187. ftanzae = — log cosa. 


188. 188, { tan*eds = tanx—«. 





n-1 
189. 189. f tan” adea= he a tan” -°ada. 
nr preset) 


190. fotne dx = log sina. 


191. f ctn’ede = — ctnaz — x. 


n-l 
192. f ctn" ade =— cnn = f etn tad, 
n— 


198. fosec ada = log tan & a >): 


194. if sec?ada = tana. 


22 TRANSCENDENTAL FUNCTIONS, 


da 
Cos” x 





195. fscotanae — 
196. foese ada = log tans. 


197. { csctada =— ctne. 








198. ( csc*xdx = (2. 
sin” x 
199. fo “— =< bit renee - sin7? b+a & COS% : 
a+b cose 2b a+b cosa 
or | ta 16g [ba @ cose + Vor ar sim 
Ve =a a+b cosa 
200. f oh ug, OOS aaa 
a+bcosx+ecsine 
si Eales 2 ed sin b ee 
Ve— Boe VP+ 2 (a+b cosa +c sing) 
1 
ee - log 
Ve +2 — a 


[ae +a(bcosx+csinxz)+ Ve +e — a’? (bsinx—ec ae 
Vb? + (a+b cosx +c sinz) 


201. f 2 sinx dx = sinx — & Cosa. 

202. {2° sinadx = 22 sina — (a — 2) cosa. 

203. f 2! sinada = (3a — 6) sinw — (a — 6%) cosa. 
204. f x” sinaxdx = — x" cosx +m f x" cosxda. 
205. [2 cosada = cosx + x sina. 

206. {2° cosa da = 2% cosa + (a — 2) sina. 


207. | a cosudx = (32° — 6) cosa + (a — 62) sina. 


TRANSCENDENTAL FUNCTIONS. 


208. fe cosxadxa = x2” sin x — ee sinxda. 
209. SB ae lees ee ft ( O8F ae 
m ee Te rate m—1 


210. eFax 2 AN Le 1 cose “1 pine g 
m—1 a) m—1 


g° Avg a" a? 
m1. { B2 dea — 32 als en 
" 3.311 5-5! 7-71 9-9! 
ont 6 


cOs@ 1 a a, 
(82 de =1 pie lh ae Oe Re Tae es 
a JS po a Se a St Hl 6-61 8.8! 


218. f sin mex sinne da = sin(m—n)x _ sin(m +n) a 
2(m— Nn) 2(m +n) 


sin(m—n)x , sin(m+n)ax 
2(m — n) 2(m +n) 


215. {sin tede= x sin-ta + V1 —2?. 














214. | cosmxcosnadz = 


216. focos tae =x costa —V1—a2*. 

217, { tanta de = «tanta —4tlog(1+2"). 

918. fetmede =x ctn'a+ slog (1 +2’). 

219. f versin tada = (# — 1) versinx + V2a—2", 

220, { (sin *e)*dz =a (sina)? —2a+2V1 —2? sina. 


oni, fz. sin ade =4[(2a°—1) sinta +avV1 — 2"). 











gee, (or sin-zdr =" Sine __1 (tide. 
. n+1 n+1J /[—# 
eee costed 8 ea de 
| a olen 
ov tan te 1 Catt tdar, 








24, fx" tan-tadzx = 
n-+1 n+1/J71+2 


24 TRANSCENDENTAL FUNCTIONS. 


295, (log sda =2 loga — x. 
226, iff Ls ba 
a 





: ; (log x)”*1. 





227. eae. log. log a. 
x log x 
dx 1 
oe pec 
ziti x (log 7)” (n — 1) (loga)""? 


log a 1 
m | d eee m-+1 i 
a9, fa og xdx = x Cue Rai ea aa 


230, iy e da =o 
a 


api. fxewde =“ (axz—1). 
a 


a” ew m 








othe | 
233. Sy egg ne 
33 fs —1 2 aa Ha | an 1 
sk Frese cia e" dat. 
a al & 
; e* (a sing — Cosa) 

. 4} e@ sing de = ———_____. 

235 f rae 


e (a cosx + sing) 
BS 8 6° GOR COG ree ee ees 
236 Jf a 


PEFINITE INTEGRALS. 25 


DEFINITE INTEGRALS, 








rth OS Wiis ; ete : Wiis 

237. ray if a>05 0, if a=0; —5, if a<o. 
) 1 1 m—1 

238. {" wtede= | [toe da= I (1). 
0 0 a 
T(n+1)=n-T(n). D2) eek) (Lee 
T(n+1)=7n!, if n is an integer T (4) = Vz. 


Cg el ay h ett een Cat) bt Tt) 
289. ("x (1—2x)"*da= Vis G@pay* ena). 


uy 5 
240. 4; sin" 2dx = i cos" a dx 
0 0 


A SS AT; (eo ee : , 
= —__________- . -, if n is an even integer. 
2.4.6...(”) 2 


st one eta ciara if m is an odd integer. 
1s Ey ey ne 





~) 
=4./7_\___“, for any value of n. 


a4. same 2, if m>0; 0, if m=0; — 4, if m<0. 
0 a 


a 


“sinxz.cosmada 
242, “SAernbe i hdl 
-: 


=0, if m<—1 orm>l; 


x if m=—1 or m=1; * if —l<m<l. 


943, ( Sinteda _ 7, 
Jo a 2 


244. ("cos (2) ett sin (2°) de = 4/5. 
0 


26 DEFINITE INTEGRALS. 


245. COSMH AL _ wom. 
0 1l+2? 2 


2c. {* oom cos ada _ ane 
ove NP 


; , 
fee 
0 4/1 — k? sin?a 





=" 14 (4) 4 (22) 4 (18) | 6 
4, A) Der Cal Ta 


= Kf, 


2 
248. f J/1 — k? sin?x. da 


3\*kt  /1.8.5\2% 
272 __ Mat pide ta 
=5|1 a Sea a 3 Ge 5 








= ff. 
249 Be eaten met nae ts 
fe etda=— fe» = P(t) 
ae ae * da = LA MS) Rema 0.) 
qrt grt 
51. [ ain p-a0? dy — 1:3-5+..(2n—1) |x 
2 ae ie a 


9 @ —2a 
252. (" Mae eats Die 
0 3 
253. f" « e-“cos mada = ———— EMER. if a>O0. 
0 a? rae 


2 


a 
: m : 
254. e-*sinmadx = ———., if a>0. 
0 m 


b2 

@ Be —- — 

Pe .@ 4a? 

255. J Ot con hed een 
0 


2 


256, (108% gy ——™. 


0 1—@w2 i \ 
257. (2824 ae 
Jo 1+2 12 


a if <1; 


DEFINITE INTEGRALLY. 


1 log a r 

258. Ed he (eer 
JS ye Man ee 
259. [" lo Lge ees am, 
l—z2 ; 4 

260. EE AW efeectc 
es a 7) a= ; 


261. ( —“- fee 
1 


Vee(;) 


262. ai a” log (“y dz= Tm+)). 
0 v 


8 | 


(m+ 1)" 
263. log sinada =. () log cosxda = — 
0 0 


264. ("a -log sinzda = — 5 1082+ 
0 


27 


28 AUXILIARY FORMULAS. 


AUXILIARY FORMULAS, 


The following formulas are sometimes useful in the reduction 
of integrals : 


265. logu=logcu-+a constant. 
266. log(—u)=logw-+ a constant. 
“sin? Vi—w © +a constant 
967. sin-tw= 4 —4gsin *(2wv?—1) +a constant. 
[ tsin2uV1—w? +a constant. 


_ tan-1 4 + a constant. 
U ; 


a | pw 
968, tan~u= as Saar 


1 — cu 


269. log(#+ yi)=4log(2’*+y") + itan*¥. 


ta +a constant. 


270. sin-tu = cos7!vV/1 — yu? = tan- Blac ap 





. PLEA | 1 
271. cos /u = sin-}/1 —v= tan*/2,— 1 = sec “-- 
U 





Lz ay 
273, sintae+sin-y=sin?(a#V1—y+yvi1 —2). 


979, tan a+ tan ‘y= tan (<* ty ) 


974. cos#+cos 'y=cos! (ay = V(1—2#)(1—7)). 





6 2 ea 
SU SLED er ee - 
275 ¥ 
276. cos“ = — 


' 977, sinwi =4$7(e* —e*)=isinha. 
978. cosxi=4(e*+e*)= cosha. 
279, log, « = (2.3025851) logia. 


Z 


Z 


_ 





C600 St Ge St im 08 PS 





OO OIG) OP 09 We 





TABLES. 


29 


The Natural Logarithms of Numbers between 1.0 and 9.9. 


1 


0.095 
0.742 
1.131 
iil 
1.629 
1.808 
1.960 
2.092 
2.208 


2 


0.182 
0.788 
1.163 
A355 


3 


0.262 
0.833 
1.194 
LA59 
1.649 | 1.668 
1.825 | 1.841 
1.974 | 1.988 
2.104 | 2.116 
2.219 | 2.230 

















0.470 | 0.531 
0.956 | 0.993 
1.281 | 1.308 
1.526 | 1.548 
1372371740 
1.887 | 1.902 
2.028 | 2.041 
2.152 | 2.163 
2.262 | 2.272 





2.282 





The Natural Logarithms of Whole Numbers from 10 to 109. 


The Values in Circular Measure of Angles which are given i 


0.0003 
0.0006 
0.0009 
0.0012 
0.0015 
0.0017 
0.0020 
0.0023 























5 


2.708 
3.219 
3.550 
3.807 
4.007 
4.174 
i517 
4475 
40% 











4.654 | 





6 





2.173 
3.258 
3.584 
3.829 
4.025 
4.190 
4.331 
4.454 
4.564 
4.663 








Degrees and Minutes. 


0.0026 
0.0029 
0.0058 
0.0087 
0.0116 
0.0145 
0.0175 
0.0349 











0.0524 
0.0698 
0.0873 |) 
0.1047 
0.1222 
0.1396 
0.1571 
0.1745 





20° 
30° 
40° 
| 50° 
60° 
70° 
80° 
B08 


0.3491 
0.5236 
0.6981 
0.8727 
1.0472 
1.2217 
1.3963 
1.5708 











100° 
110° 
120° 
1309 
140° 
150° 
160° 
170° 








1.7453 
1.9199 
2.0944 
2.2689 
2.4435 
2.6180 
2.7925 
2.9671 





30 





TABLES. 


NATURAL TRIGONOMETRIC FUNCTIONS. 







































































Angle. Sin. Csc. Tan. Ctn. Sec.-./'" ¥@on: 
0° 0.000 oe) 0.000 a) 1.000 1.000 
1 || 0.017 | 57.30 DOL 4) S724 1.000 1.000 
2 Up .0.035) 928-65 0.035 | 28.64 1.001 0.999 
3 WeOO0S20 019511 0.052 | 19.08 1.001 0.999 
+ | 0.070 | 14.34 0.070 | 14.30 1.002 0.998 
5° || ~(0.087 | 11.47 OST tal ARS 1.004 0.996 
6 0.105 9.567 0.105 9.55 1.006 | 0.995 
7 0.122 8.206 0.123 8.144 1.008 0.993 
8 0.139 7.185 0.141 7.115 1.010 0.990 
9 0.156 6.392 0.158 6.314 1.012 0.988 

10° 0.174 57g 0.176 5.671 1.015 0.985 
11 0.191 5.241 0.194 5.145 1.019 0.982 
12 0.208 4.810 0.213 4.705 1.022 0.978 
13 |} 0.225 4.445 0.231 4.331 1.026 0.974 
14 0.242 4.134 0.249 4.011 1.031 0.970 
15° 0.259 3.864 0.268 3.732 1.035 0.966 
16 0.276 3.628 0.287 3.487 1.040 0.961 
17 0.292 3.420 0.306 3.271 1.046 0.956 
18 0.309 3:2505 1 -<O.325 3.078 1.051 0.951 
19 0.326 3.072 0.344 2.904 1.058 0.946 
20° 0.342 | 2.924 | 0.364 | 2.747 | 1.064 | 0.940 
21 0.358 | 2.790 | 0.384 | 2.605 | 1.071 | 0.934 
22 0.375 | 2.669 | 0.404 | 2475 | 1.079 | 0.927 
23 0.391 | 2.559 | 0.424 | 2.356 | 1.086 | 0.921 
24 0.407 | 2.459 | 0.445 | 2.246 | 1.095 | 0.914 
25° 0.423 2.366 0.466 2145 1.103 0.906 
26 0.438 2.281 0.488 2.050 1.113 0.899 
27 0.454 2.203 0.510 1.963 1.122 0.891 
28 0.469 2.130 0.532 1.881 1.133 0.883 
29 0.485 2.063 0.554 1.804 1.143 0.875 
30° 0.500 2.000 0.577 1.732 2. Lae 0.866 
31 OS1S 142 0.601 1.664 1.167 0.857 
32 0.530 1.887 0.625 1.600 1,179 0.848 
33 0.545 1.836 0.649 1.540 1.192 0.839 
34 0.559 1.788 0.675 1.483 1.206 0.829 
35° 0.574 1.743 0.700 1.428 bee 0.819 
36 0.588 1.701 0.727 1.376 1.236 | 0.809 
37 0.602 1.662 0.754 1.327 1.252 0.799 
38 0.616 1.624 0.781 1.280 1.269 0.788 
39 0.629 1.589 0.810 1.235 1.287 0.777 
40° 0.643 1.556 0.839 1,192 1.305 0.766 
41 0.656 Let 0.869 1.150 13325 0.755 
42 0.669 1494 0.900 irae! 1.346 0.743 
43 0.682 1.466 | 0.933 1.072 1.367 0.731 
44 0.695 1.440 0.966 1.036 1.390 0.719 
45° 0.707 1.414 1.000 1.000 14140) 078s 


| | | | 

















TABLES. 


31 


Values of the Complete Elliptic Integrals, A and /, for Different 
Values of the Modulus, xk. 

















E |lsin-2k 
1.5708 || 30° 
1.5707 || 31° 
1.5703 || 32° 
1.5697 || 33° 
1.5689 || 34° 
1.5678 || 35° 
1.5665 || 36° 
1.5649 || 37° 
1.5632 || 38° 
1.5611 |} 39° 
1.5589 || 40° 
1.5564 || 41° 
1.5537 || 42° 
1.5507 || 43° 
1.5476 || 44° 
1.5442 || 45° 
1.5405 || 46° 
1.5367 || 47° 
1.5326 || 48° 
1.5283 || 49° 
1.5238 || 50° 
1.5191 || 51° 
1.5141 || 52° 
1.5090 || 53° 
1.5037 || 54° 
1.4981 || 55° 
1.4924 || 56° 
1.4864 || 57° 
1.4803 || 58° 
1.4740 || 59° 








1.6858 
1.6941 
1.7028 
Lary 
1.7214 
1.7312 
1.7415 
1.7522 
1.7633 
1.7748 
1.7868 
1.7992 
1.8122 
1.8256 
1.8396 
1.8541 
1.8691 
1.8848 
1.9011 
1.9180 
1.9356 
10539 
1.9729 
1.9927 
2.0133 
2.0347 
2.0571 
2.0804 
2.1047 
2.1300 











isin 1k 


| 60° 
61° 
id 
os” 
64° 





























32 


TABLES, 


The Common Logarithms of [(n) for Values of n between 1 and 2. 


——_—__—_ 


1.01 
1.02 
1.03 
1.04 
1.05 
1.06 
1.07 
1.08 
1.09 
1.10 
1.11 
1.12 
1.13 
1.14 
1.15 
116 
1.17 
1.18 
1.19 








logy) '(n) 





1.9975 
1.9951 
1.9928 
1.9905 
1.9883 
1.9862 
1.9841 
1.9821 
1.9802 
1.9783 
1.9765 
1.9748 
1.9731 
1.9715 
1.9699 
1.9684 
1.9669 
1.9655 
1.9642 














1.21 
1°22 
1.23 
1.24. 
V5 
1.26 
127 
1.28 
1.29 
1.30 
154 
1.32 
1.33 
LO+ 
i635 
1.36 
MEY 
1.38 
Ls 





logy) T(n) 





1.9617 
1.9605 
1.9594 
1.9583 





1.9573 
1.9564 
1.9554 
1.9546 
1.9538 
1.9530 
1.9523 
1.9516 
1.9510 | 
1.9505 
1.9500 
1.9495 
1.9491 
1.9487 
1.9483 











1.41 
1.42 
1.43 
1.44 





logy) T'(n) 





1.9478 
1.9476 
1.9475 
1.9473 


1.45 11.9473 


1.46 
1.47 
1.48 
Tay. 
1.50 


1.9472 
1.9473 
1.9473 
1.9474 
1.9475 


1.51 |1.9477 


1.52 
1.53 
Pt 
1.55 
1.56 
1.57 
1.58 
1.59 


1.9479 
1.9482 
1.9485 
1.9488 
1.9492 


1.9496 || 


1.9501 
1.9506 


1.20 11.9629 || 1.40 11.9481 || 1.60 [1.9511 











n 





1.61 
1.62 
1.63 
1.64 
1.65 
1.66 
1.67 
1.68 
1.69 
1.70 
Ae 
1.72 
ey 
1.74 
1.75 
1.76 
1.77 
1.78 
1.79 
1.80 


logy) T'(n) 





1.9517 
1.9523 








1.9529 
1.9536 
1.9543 
1.9550 
1.9558 
1.9566 
1.9575 
1.9584 
1.9593 
1.9603 
1.9613 
1.9623 
1.9633 
1.9644 
1.9656 
1.9667 
1.9679 


1.9691 





1.81 
1.82 
1.83 
1.84 
1.85 
1.86 
1.87 
1.88 
1.89 
1.90 
1.91 
1.92 
1.93 
LE 
LS 
1.96 
1.97 
1.98 
1.99 


logy) T(r) 





1.9704 
1.9717 
1.9730 
1.9743 
1.9757 
1.9771 
1.9786 
1.9800 
1.9815 
1.9831 
1.9846 
1.9862 
1.9878 
1.9895 
19912 
1.9929 
1.9946 
1.9964 
1.9982 


2.00 |0.0000 





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ELEMENTS OF THE INTEGRAL 


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